Solution

First women choose the chairs from among 1 to 4 chairs. i.e., total number of chairs is 4 . Since, there are two women, so number of arrangements $={ }^{4} P_2$ ways.

Now, men have to choose chairs from remaining 6 chairs.

Since, there are 3 men, so number can be arranged in ${ }^{6} P_3$ ways.

$\therefore$ Total number of possible arrangements $={ }^{4} P_2 \times{ }^{6} P_3$

$ \begin{aligned} & =\frac{4 !}{4-2 !} \times \frac{6 !}{6-3 !} \\ & =\frac{4 !}{2 !} \times \frac{6 !}{3 !} \\ & =\frac{4 \times 3 \times 2 !}{2 !} \times \frac{6 \times 5 \times 4 \times 3 !}{3 !} \\ & =4 \times 3 \times 6 \times 5 \times 4=1440 \end{aligned} $

Solution

The letters of the word ‘RACHIT’ in alphabetical order are A, C, H, I, R and T.

Now, $\quad$ words beginning with $A=5$ !

words beginning with $C=5$ !

words beginning with $H=5$ !

words beginning with $I=5$ !

Word beginning with R i.e., RACHIT $=1$

$\therefore \quad$ Rank of the word ‘RACHIT’ in dictionary $=4 \times 5$ ! $+1=4 \times 120+1$

$ =480+1=481 $

Solution

Since, candidate cannot attempt more than 5 questions from either group.

Thus, he is able to attempt minimum two questions from either group.

The number of questions attempted from each group is given in following table

Since, each group have 6 questions and total attempted 7 questions.

$\therefore \quad$ Total number of possible ways $={ }^{6} C_5 \times{ }^{6} C_2+{ }^{6} C_4 \times{ }^{6} C_3+{ }^{6} C_3 \times{ }^{6} C_4+{ }^{6} C_2 \times{ }^{6} C_5$

$ \begin{aligned} & =2[{ }^{6} C_5 \times{ }^{6} C_2+{ }^{6} C_4 \times{ }^{6} C_3] \\ & =2[6 \times 15+15 \times 20] \\ & =2[90+300] \\ & =2 \times 390=780 \end{aligned} $

Solution

Total number of points $=18$

Out of which 5 points are collinear, we get a straight line by joining any two points.

$\therefore \quad$ Number of straight line formed by joining the 18 points taking 2 at a time $={ }^{18} C_2$

and number of straight line formed by joining 5 points taking 2 at a time $={ }^{5} C_2$

But 5 collinear points, when joined pairwise give only one line.

$\therefore \quad$ Required number of straight line $={ }^{18} C_2-{ }^{5} C_2+1$

$ =153-10+1=144 $

Solution

Total number of person $=8$

Number of person to be selected $=6$

It is given that, if $A$ is chosen then, $B$ must be chosen.

Therefore, following cases arise.

Case I When $A$ is chosen, $B$ must be chosen

Number of ways $={ }^{8-2} C_{6-2}={ }^{6} C_4$

Case II When $A$ is not chosen.

Then, $B$ may be chosen.

$\therefore \quad$ Number of ways $={ }^{8-1} C_6={ }^{7} C_6$

Hence, required number of ways $={ }^{6} C_4+{ }^{7} C_6$

$ =15+7=22 $

Solution

$\because \quad$ Total number of persons $=12$

and number of persons to be selected $=5$

Out of 12 persons a chairperson is selected $={ }^{12} C_1=12$ ways

Now, remaining 4 persons are selected out of 11 persons.

$\therefore$ Number of ways $={ }^{11} C_4=330$

$\therefore$ Total number of ways to form a committee of 5 persons $=12 \times 330=3960$

Solution

There are 26 English alphabets and 10 digits (0 to 9).

Since, it is given that each plate contains two different letters followed by three different digits.

$\therefore \quad$ Arrangement of 26 letters, taken 2 at a time $={ }^{26} P_2=\frac{26 !}{24 !}=26 \times 25=650$ and three-digit number can be formed out of the 10 digits $={ }^{10} P_3=10 \times 9 \times 8=720$ ways

$\therefore$ Total number of licence plates $=650 \times 720=468000$

Solution

It is given that bag contains 5 black and 6 red balls.

So, 2 black balls is selected from 5 black balls in ${ }^{5} C_2$ ways.

and 3 red balls are selected from 6 red balls in ${ }^{6} C_3$ ways.

$\therefore$ Total number of ways in which 2 black and 3 red balls are selected $={ }^{5} C_2 \times{ }^{6} C_3$

$=10 \times 20=200$ ways

Solution

Total number of things $=n$

We have to arrange $r$ things out of $n$ in which three things must occur together.

Therefore, combination of $n$ things taken $r$ at a time in which 3 things always occurs

$ ={ }^{n-3} C_{r-3} $

If three things taken together, then it is considered as 1 group.

Arrangement of these three things $=3$ !

Now, $\quad$ we have to arrange $=r-3+1=(r-2)$ objects

$\therefore \quad$ Arranged of $(r-2)$ objects $=r-2$ !

$\therefore$ Total number of arrangements $={ }^{n-3} C_{r-3} \times r-2 ! \times 3$ !

Solution

Number of letters in the word ‘TRIANGLE’ $=8$, out of which 5 are consonants and 3 are vowels.

If vowels are not together, then we have following arrangement.

Consonants can be arranged in $=5 !=120$ ways and vowels can occupy at 6 places.

The 3 vowels can be arranged at 6 place in ${ }^{6} P_3$ ways $=\frac{6 !}{6-3 !}=\frac{6 !}{3 !}$

$ =\frac{6 \times 5 \times 4 \times 3 !}{3 !}=120 $

Total number of arrangement $=120 \times 120=14400$

Solution

We know that a number is divisible by 5 , If at the units place of the number is 0 or 5 .

We have to form 4 -digit number which is greater than 6000 and less than 7000 . So, unit digit can be filled in 2 ways.

Since, repeatition is not allowed. Therefore, tens place can be filled in 7 ways, similarily hundreds place can be filled in 8 ways.

But we have to form a number greater than 6000 and less than 7000 .

Hence, thousand place can be filled in only 1 ways.

Total number of integers $=1 \times 8 \times 7 \times 2$

$ =14 \times 8=112 $

Solution

Given that, $P_1, P_2, \ldots, P_{10}$, are 10 persons, out of which 5 persons are to be arranged but $P_1$ must occur whereas $P_4$ and $P_5$ never occur.

$\therefore$ Selection depends on only $10-3=7$ persons

As, we have already occur $P_1$, Therefore, we have to select only 4 persons out of 7 .

Number of selection $={ }^{7} C_4=\frac{7 !}{4 !(7-4) !}=\frac{7 !}{4 ! 3 !}=\frac{5040}{24 \times 6}=35$

$\therefore$ Required number of arrangement of 5 persons $=35 \times 5 !=35 \times 120=4200$

Solution

Total number of ways $={ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3+{ }^{10} C_4+{ }^{10} C_5+{ }^{10} C_6+\ldots+{ }^{10} C_{10}$

$ \begin{aligned} & =2^{10}-1 \quad[\because{ }^{n} C_0+{ }^{n} C_1+{ }^{n} C_2+\ldots=2^{n}] \\ & =1024-1=1023 \end{aligned} $

Solution

There are 2 white, three black and four red balls.

We have to draw 3 balls, out of these 9 balls in which atleast one black ball is included.

Hence, we can select the balls in the following ways.

$\therefore$ Required number of selections $={ }^{3} C_1 \times{ }^{6} C_2+{ }^{3} C_2 \times{ }^{6} C_1+{ }^{3} C_3 \times{ }^{6} C_0$

$ \begin{aligned} & =3 \times 15+3 \times 6+1 \\ & =45+18+1=64 \end{aligned} $

Solution

Given, ${ }^n C_{r-1} = 36 \quad \quad …(i)$

$\Rightarrow { }^n C_r = 84 \quad \quad …(ii)$

$\Rightarrow { }^n C_{r-1} = 126 \quad \quad …(iii)$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}} =\frac{36}{84} \\ $

$\Rightarrow \frac{n !}{(r-1) !{n-(r-1)} !} \cdot \frac{r !(n-r) !}{n !} =\frac{3}{7} \\ $

$\Rightarrow \frac{1}{(r-1) !(n-r+1) !} \cdot \frac{r(r-1) !(n-r) !}{1} =\frac{3}{7} \\ $

$\Rightarrow \frac{1 \cdot r}{(n-r+1)(n-r) !} \cdot(n-r) ! =\frac{3}{7} \Rightarrow \frac{r}{n-r+1}=\frac{3}{7} \\ $

$\Rightarrow 10 r-3 n = $

$\because{ }^{n} C_{r}=\frac{n !}{(n-r ! r !)}$ and $n !=n(n-1) !$

On dividing Eq. (ii) by Eq. (iii), we get

$ \begin{aligned} & \Rightarrow \quad \frac{{ }^{n} C_{r}}{{ }^{n} C_{r}+1}=\frac{84}{126} \\ & \Rightarrow \quad \frac{1}{r !(n-r) !} \cdot \frac{(r+1) !(n-r-1) !}{n !}=\frac{14}{21} \\ & \Rightarrow \quad \frac{(r+1) r !(n-r-1) !}{r}=\frac{2}{3} \Rightarrow \frac{r+1}{n-r}=\frac{2}{3} \\ & 3 r+3=2 n-2 r \Rightarrow 2 n-5 r=3 \\ & \text { On multiplying Eq. (iv) by 2 and Eq. (v) by 3, we get } \\ & 20 r-6 n=6 \\ & 6 n-15 r=9 \end{aligned} $

On adding Eqs. (vi) and (vii),

$ \begin{matrix} \text { From Eq. (v), } & 2 n=3+15 \\ \Rightarrow & 2 n=18 \Rightarrow n=9 \\ \therefore & { }^{r} C_2={ }^{3} C_2=\frac{3 !}{2 ! 1 !}=\frac{3 \times 2 !}{2 !}=3 \end{matrix} $

$ 5 r=15 \Rightarrow r=3 $

Solution

Here, we have to find the number of integers greater than 7000 with the digits $3,5,7,8$ and 9 . So, with these digits we can make maximum five-digit number because repeatition is not allowed.

Now, all the five-digit numbers are greater than 7000 .

Number of ways of forming 5-digit number $=5 \times 4 \times 3 \times 2 \times 1=120$

and all the four-digit numbers greater than 7000 can be formed in following manner.

Thousand place can be filled in 3 ways. Hundred place can be filled in 4 ways. Tenth place can be filled in 3 ways. Units place can be filled in 2 ways.

Thus, we have total number of 4-digit number $=3 \times 4 \times 3 \times 2=72$

$\therefore$ Total number of integers $=120+72=192$

Solution

It is given that no two lines are parallel means all line are intersecting and no three lines are concurrent means three lines intersect at a point.

Since, we know that for one point of intersection, we required two lines.

$\therefore \quad$ Number of point of intersection $={ }^{20} C_2=\frac{20 !}{2 ! 18 !}=\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}$

$ =\frac{20 \times 19}{2}=19 \times 10=190 $

Solution

If first two digit is 41 , the remaining 4 digits can be arranged in

$ \begin{aligned} & ={ }^{8} P_4=\frac{8 !}{8-4 !}=\frac{8 !}{4 !} \\ & =\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !} \\ & =8 \times 7 \times 6 \times 5=1680 \end{aligned} $

Similarly, if first two digit is $42,46,62$, or 64 , the remaining 4 digits can be arranged in ${ }^{8} P_4$ ways i.e., 1680 ways.

$\therefore$ Total number of telephone numbers have all six digits distinct $=5 \times 1680=8400$

Solution

It is given that 2 questions are compulsory out of 5 questions.

So, these two questions are always included in the selection.

We know that, the selection of $n$ distinct objects taken $r$ at a time in which $p$ objects are always included in ${ }^{n-p} C_{r-p}$ ways.

$\therefore \quad$ Total number of ways $={ }^{5-2} C_{4-2}={ }^{3} C_2$

$ =\frac{3 !}{2 ! 1 !}=\frac{3 \times 2 !}{2 !}=3 $

Solution

Let the convex polygon has $n$ sides.

$\therefore \quad$ Number of diagonals $={ }^{n} C_2-n$

According to the question,

${ }^{n} C_2-n =44 \\ $

$\frac{n!}{2!(n-2)!} - n = 44$

$\Rightarrow \frac{n(n-1)!}{2}-n =44 $

$\Rightarrow n \Big[\frac{n-1}{2} - 1\Big] = 44\\ $

$\Rrightarrow n\Big (\frac{n-1-2}{2} \Big) = 44$

$\Rightarrow n(n-3) =44 \times 2 \Rightarrow n(n-3)=88 \\ $

$\Rightarrow n^{2}-3 n-88 =0 \quad \Rightarrow(n-11)(n+8)=0 \\ $

$\Rightarrow n =11,-8 \quad \quad [\because n \neq -8] $

$\therefore n =11 $

Solution

It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.

$\therefore \quad$ Required arrangements $=\frac{\text { Total arrangement }}{\text { Equally likely arrangement }}=\frac{18 !}{6 ! 6 ! 6 !}$

Solution

Total number of marbles $=6$ white +5 red $=11$ marbles

(i) If they can be of any colour means we have to select 4 marbles out of 11.

$\therefore$ Required number of ways $={ }^{11} C_4$

(ii) If two must be white, then selection will be ${ }^{6} C_2$ and two must be red, then selection will be ${ }^{5} C_2$.

$\therefore$ Required number of ways $={ }^{6} C_2 \times{ }^{5} C_2$

(iii) If they all must be of same colour, then selection of 4 white marbles out of $6={ }^{6} C_4$

and selection of 4 red marble out of $5={ }^{5} C_4$

$\therefore \quad$ Required number of ways $={ }^{6} C_4+{ }^{5} C_4$

Solution

Total number of players $=16$

We have to select a team of 11 players

(i) include 2 particular players $={ }^{16-2} C_{11-2}={ }^{14} C_9$

[since, selection of $n$ objects taken $r$ at a time in which $p$ objects are always included is ${ }^{n-p} C_{r-p}$ ]

(ii) Exclude 2 particular players $={ }^{16-2} C_{11}={ }^{14} C_{11}$

[since, selection of $n$ objects taken $r$ at a time in which $p$ objects are never included

is ${ }^{n-p} C_{r}$ ]

Solution

Total students in each class $=20$

We have to selects atleast 5 students from each class.

Hence, selection of sport team of 11 students from each class is given in following table

$\therefore$ Total number of ways of selecting a team of 11 players $={ }^{20} C_5 \times{ }^{20} C_6+{ }^{20} C_6 \times{ }^{20} C_5$

$ =2 \times{ }^{20} C_5 \times{ }^{20} C_6 $

Solution

Number of girls $=4$ and Number of boys $=7$

We have to select a team of 5 members provided that

(i) team having no girls.

$\therefore \quad$ Required selection $={ }^{7} C_5=\frac{7 !}{5 ! 2 !}=\frac{7 \times 6}{2}=21$

(ii) atleast one boy and one girl

$\therefore \quad$ Required selection $={ }^{7} C_1 \times{ }^{4} C_4+{ }^{7} C_2 \times{ }^{4} C_3+{ }^{7} C_3 \times{ }^{4} C_2+{ }^{7} C_4 \times{ }^{4} C_1$

$ \begin{aligned} & =7 \times 1+21 \times 4+35 \times 6+35 \times 4 \\ & =7+84+210+140=441 \end{aligned} $

(iii) when atleast three girls are included $={ }^{4} C_3 \times{ }^{7} C_2+{ }^{4} C_4 \times{ }^{7} C_1$

$ =4 \times 21+7=84+7=91 $

Solution

$\because \quad$ Total number of $men=10$

and total number of women $=7$

We have to form a committee containing atleast 3 men and 2 women.

Number of ways $={ }^{10} C_3 \times C_3+{ }^{10} C_4 \times C_2$

If two particular women to be always there

$\therefore \quad$ Number of ways $={ }^{10} C_4 \times{ }^{5} C_0+{ }^{10} C_3 \times{ }^{5} C_1$

Total number of committee when two particular women are never together

$ \begin{aligned} & =\text { Total }- \text { Together } \\ & =({ }^{10} C_3 \times{ }^{7} C_3+{ }^{10} C_4 \times{ }^{7} C_2)-({ }^{10} C_4 \times{ }^{5} C_0+{ }^{10} C_3 \times{ }^{5} C_1) \\ & =(120 \times 35+210 \times 21)-(210+120 \times 5) \\ & =4200+4410-(210+600) \\ & =8610-810=7800 \end{aligned} $

Solution

(a) Given that,

$ { }^{n} C_{12}={ }^{n} C_8 $

$ \begin{aligned} & \Rightarrow \quad{ }^{n} C_{n-12}={ }^{n} C_8 \\ & \Rightarrow \quad n-12=8 \quad \quad {[\because{ }^{n} C_{r}={ }^{n} C_{n-r}]}\\ & \Rightarrow \quad n=12+8=20 \\ \end{aligned} $

Solution

(b) Number of outcomes when tossing a coin 1 times $=2$ (head or tail)

$\therefore$ Total possible outcomes when a coin tossed 6 times $=2^{6}=64 \quad \quad [\because 2^{n}$ or n time tossed coin]

Solution

(c) Given, digits 2, 3, 4 and 7, we have to form four-digit numbers using these digits.

$\therefore \quad$ Required number of ways $={ }^{4} P_4=4 !=4 \times 3 \times 2 !=24$

Solution

(b) If we fixed 3 at units place.

Total possible number is 3 ! i.e., 6 .

Sum of the digits in unit place of all these numbers $=3 ! \times 3$

Similarly, if we fixed 4,5 and 6 at units place, in each case total possible numbers are 3 !.

Required sum of unit digits of all such numbers $=(3+4+5+6) \times 3$ !

$ =18 \times 3 !=18 \times 6=108 $

Solution

(c) Given that, total number of vowels $=4$

and total number of consonants $=5$

Total number of words formed by 2 vowels and 3 consonants

$ \begin{aligned} & ={ }^{4} C_2 \times{ }^{5} C_3=\frac{4 !}{2 ! 2 !} \times \frac{5 !}{3 ! 2 !} \\ & =\frac{4 \times 3 !}{2 ! 2 !} \times \frac{5 \times 4 \times 3 \times 2 !}{3 ! \times 2 !}=\frac{4 \times 5 \times 4 \times 3}{4} \\ & =5 \times 4 \times 3=60 \end{aligned} $

Choose what order they appear in 5! i.e., 120.

So, total number of words $=60 \times 120=7200$

Solution

(a) We know that, a number is divisible by 3 , when sum of digits in the number must be divisible by 3 .

So, if we consider the digits $0,1,2,4,5$, then $(0+1+2+4+5)=12$

We see that, sum is divisible by 3 . Therefore, five-digit numbers using the digit

$0,1,2,4,5=4 \times 4 \times 3 \times 2 \times 1=96$

and if we consider the digit $1,2,3,4,5$, then $(1+2+3+4+5=15)$

This sum is also divisible by 3 .

So, five-digit number can be formed using the digit 1, 2, 3, 4, 5 in 5 ! ways.

Total number of ways $=96+5 !=96+120=216$

Solution

(b) Let the total number of person in the room is $n$. We know that, two person form 1 hand shaken.

$\therefore \quad$ Required number of hand shakes $={ }^{n} C_2=\frac{n !}{2 !(n-2) !}=\frac{n(n-1)}{2}$

According to the question, $\frac{n(n-1)}{2}=66$

$\Rightarrow \quad n(n-1)=132$

$\Rightarrow \quad n^{2}-n-132=0$

$\Rightarrow \quad(n-12)(n+11)=0$

$\Rightarrow \quad n=12,-11 \quad$ [inadmissible]

$\therefore \quad n=12$

Solution

(d) Total number of triangles formed from 12 points taking 3 at a time $={ }^{12} C_3$

But out of 12 points 7 are collinear. So, these 7 points constitute a straight line mean no triangle is formed by joining these 7 points.

$\therefore \quad$ Required number of triangles $={ }^{12} C_3-{ }^{7} C_3=220-35=185$

Solution

(b) To form parallelogram we required a pair of line from a set of 4 lines and another pair of line from another set of 3 lines.

$\therefore$ Required number of parallelograms $={ }^{4} C_2 \times{ }^{3} C_2=6 \times 3=18$

Solution

(c) Total number of players $=22$

We have to select a team of 11 players. Selection of 11 players when 2 of them is always included and 4 are never included.

Total number of players $=22-2-4=16$

$\therefore$ Required number of selections $={ }^{16} C_9$

Solution

(d) If all the digits repeated, then number of 5 digit telephone numbers can be formed in $10^{5}$ ways and if no digit repeated, then 5 -digit telephone numbers can be formed in ${ }^{10} P_5$ ways.

$\therefore$ Required number of ways $=10^{5}-{ }^{10} P_5=100000-\frac{10 !}{5 !}$

$ \begin{aligned} & =100000-10 \times 9 \times 8 \times 7 \times 6 \\ & =100000-30240=69760 \end{aligned} $

Solution

(a) $\because$ Number of men $=4$

and number of women $=6$

It is given that committee includes two men and exactly twice as many women as men.

Thus, possible selection is given in following table

Required number of committee formed $={ }^{4} C_2 \times{ }^{6} C_4+{ }^{4} C_3 \times{ }^{6} C_6$

$ =6 \times 15+4 \times 1=94 $

Solution

(c) We have to form 9-digit numbers with the digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 cannot be placed at the first place from left. So, first place from left can be filled in 9 ways. Since, repetition is not allowed, so remaining 8 places can be filled in 9 ! ways.

$\therefore$ Required number of ways $=9 \times 9$ !

Solution

(b) Total number of letters in the word article is 7, out of which A, E, I are vowels and R, T, $C, L$ are consonants.

Since, it is given that vowels occupy even place, therefore the arrangement of vowel, consonant can be understand with the help of following diagram.

Now, vowels can be placed at 2, 4 and 6 th position.

Therefore, number of arrangement $={ }^{3} P_3=3$ ! $=6$ ways

and consonants can be placed at 1, 3,5 and 7th position.

Therefore, number of arrangement $={ }^{4} P_4=4$ ! $=24$

$\therefore$ Total number of words $=6 \times 24=144$

Solution

(b) Possible number of choosing green dyes $=2^{5}$

Possible number of choosing blue dyes $=2^{4}$

Possible number of choosing red dyes $=2^{3}$

If atleast one blue and one green dyes are selected.

Then, total number of selection $=(2^{5}-1)(2^{4}-1) \times 2^{3}=3720$

Solution

Given that, ${ }^{n} P_{r}=840$ and ${ }^{n} C_{r}=35$

$\because{ }^{n} P_{r}={ }^{n} C_{r} \cdot r$ !

$ \begin{matrix} \Rightarrow & 840=35 \times r ! \\ \Rightarrow & r !=\frac{840}{35}=24 \\ \Rightarrow & r !=4 \times 3 \times 2 \times 1 \\ \Rightarrow & r !=4 ! \\ \therefore & r=4 \end{matrix} $

Solution

${ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7={ }^{15} C_{15-8}+{ }^{15} C_{15-9}-{ }^{15} C_6-{ }^{15} C_7 \quad[\because{ }^{n} C_{r}={ }^{n} C_{n-r}]$

$ ={ }^{15} C_7+{ }^{15} C_6-{ }^{15} C_6-{ }^{15} C_7 $

$=0$

Solution

Number of permutations of $n$ different things taken $r$ at a time when repetition is allowed $=n^{r}$

Solution

Total number of letters in the word ‘INTERMEDIATE’ $=12$ out of which 6 are consonants and 6 are vowels. The arrangement of these 12 alphabets in which two vowels never come together can be understand with the help of follow manner.

6 consonants out of which 2 are alike can be placed in $\frac{6 !}{2 !}$ ways and 6 vowels, out of which 3 E’s alike and 2 I’s are alike can be arranged at seven place in ${ }^{7} P_6 \times \frac{1}{3 !} \times \frac{1}{2 !}$ ways.

$\therefore$ Total number of words $=\frac{6 !}{2 !} \times{ }^{7} P_6 \times \frac{1}{3 !} \times \frac{1}{2 !}=151200$

Solution

Required number of ways $={ }^{5} C_2 \times{ }^{7} C_1+{ }^{5} C_3$

[since, at least two red]

$ \begin{aligned} & =10 \times 7+10 \\ & =70+10=80 \end{aligned} $

Solution

Among the digits $0,1,2,3,4,5,6,7,8,9$, clearly $1,3,5,7$ and 9 are odd.

$\therefore \quad$ Number of six-digit numbers $=5 \times 5 \times 5 \times 5 \times 5 \times 5=5^{6}$

Solution

Let the number of team participating in championship be $n$.

Since, it is given that every two teams played one match with each other.

$\therefore$ Total match played $={ }^{n} C_2$

According to the question,

${ }^{n} C_2 =153 \\ $

$\Rightarrow \frac{n(n-1)}{2} =153 \\ $

$\Rightarrow n^{2}-n =306 \\ $

$\Rightarrow n^{2}-n-306 =0 \\ $

$\Rightarrow (n-18)(n+17) =0 \\ $

$\Rightarrow n =18,-17 \\ $

$\therefore n =18 $

Solution

The arrangement can be understand with the help of following figure.

Thus, ’ + ’ sign can be arranged in 1 way because all are identical. and 4 negative signs can be arranged at 7 places in ${ }^{7} C_4$ ways.

$\therefore \quad$ total number of ways $={ }^{7} C_4 \times 1$

$ \begin{aligned} & =\frac{7 !}{4 ! 3 !}=\frac{7 \times 6 \times 5 \times 4 !}{3 ! \times 4 !} \\ & =\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \text { ways } \end{aligned} $

Solution

Since, there are 2 white, 3 black and 4 red balls. It is given that atleast one black ball is to be included in the draw.

$\therefore$ Required number of ways $={ }^{3} C_1 \times{ }^{6} C_2+{ }^{3} C_2 \times{ }^{6} C_1+{ }^{3} C_3$

$ \begin{aligned} & =3 \times 15+3 \times 6+1 \\ & =45+18+1=64 \end{aligned} $

Solution

False

Required number of lines $={ }^{12} C_2-{ }^{5} C_2+1$

Solution

False

Required number of ways $=5^{3}=125$

Solution

False

Arrangement of $n$ things, taken $r$ at a time in which $m$ things occur together, we considered these $m$ things as 1 group.

Number of object excluding those $m$ objects $=(r-m)$

Now, first we have to arrange $(r-m+1)$ objects.

Number of arrangements $=(r-m+1)$ ! and $m$ objects which we consider as 1 group, can be arranged in $m$ ! ways.

$\therefore \quad$ Required number of arrangements $=(r-m+1) ! \times m$ !

Solution

True

There are three types of animals and stalls available for 12 animals.

Number of ways of loading $=3^{12}$

Solution

True

If some or all objects taken at a time, then number of selection would be

$ { }^{n} C_1+{ }^{n} C_2+{ }^{n} C_3+\ldots+{ }^{n} C_{n}=2^{n}-1 \quad[\because{ }^{n} C_0+{ }^{n} C_1+{ }^{n} C_2+\ldots+{ }^{n} C_{n}=2^{n}] $

Solution

Total number of selection $=[(4+1)(5+1)-1]-5$

$ \begin{aligned} & =(5 \times 6-1)-5 \\ & =(30-1)-5=24 \end{aligned} $

Solution

True

After seating 4 on one side and 3 on the other side, we have to select out of $11 ; 5$ on one side and 6 on the other side.

Now, remaining selecting of one half side $={ }^{(18-4-3)} C_5={ }^{11} C_5$

and $\quad$ the other half side $={ }^{(11-5)} C_6={ }^{6} C_6$

Total arrangements $={ }^{11} C_5 \times 9 ! \times{ }^{6} C_6 \times 9 !$

$ \begin{aligned} & =\frac{11 !}{5 ! 6 !} \times 9 ! \times 1 \times 9 ! \\ & =\frac{11 !}{5 ! 6 !} \times 9 ! \times 9 ! \end{aligned} $

Solution

False

He can attempt questions in following manner

Number of ways of attempting 7 questions

$ \begin{aligned} & ={ }^{6} C_2 \times{ }^{6} C_5+{ }^{6} C_3 \times{ }^{6} C_4+{ }^{6} C_4 \times{ }^{6} C_3+{ }^{6} C_5 \times{ }^{6} C_2 \\ & =2({ }^{6} C_2 \times{ }^{6} C_5+{ }^{6} C_3 \times{ }^{6} C_4) \\ & =2(15 \times 6+20 \times 15) \\ & =2(90+300) \\ & =2 \times 390=780 \end{aligned} $

Solution

False

We have to select 3 scheduled caste candidate out of 5 in ${ }^{5} C_3$ ways.

and we have to select 9 other candidates out of 22 in ${ }^{22} C_9$ ways.

$\therefore$ Total number of selections $={ }^{5} C_3 \times{ }^{22} C_9$

Matching The Columns

Solution

There are three books of Mathematics 4 of Physics and 5 on English.

(i) One book of each subject $={ }^{3} C_1 \times{ }^{4} C_1 \times{ }^{5} C_1$

$ =3 \times 4 \times 5=60 $

(ii) Atleast one book of each subject $=(2^{3}-1) \times(2^{4}-1) \times(2^{5}-1)$

$ =7 \times 15 \times 31=3255 $

(iii) Atleast one book of English =Selection based on following manner

Solution

(i) Boys and girls alternate

Total arrangements $=(5 !)^{2}+(5 !)^{2}$

(ii) No two girls sit together $=5 ! 6$ !

(iii) All the girls sit together $=2 ! 5 ! 5$ !

(iv) All the girls are never together $=10 !-5 ! 6$ !

Solution

(i) We have to select 2 professors out of 10 and 3 lecturers out of $20={ }^{10} C_2 \times{ }^{20} C_3$

(ii) When a particular professor included $={ }^{10}{ }^{-1} C_1 \times{ }^{20} C_3={ }^{9} C_1 \times{ }^{20} C_3$

(iii) When a particular lecturer included $={ }^{10} C_2 \times{ }^{19} C_2$

(iv) When a particular lecturer excluded $={ }^{10} C_2 \times{ }^{19} C_3$

Solution

(i) Total numbers of 4 digit formed with digits 1, 2, 3, 4, 5, 6, 7

$ =7 \times 6 \times 5 \times 4=840 $

(ii) When a number is divisible by 2. At its unit place only even numbers occurs. Total numbers $=4 \times 5 \times 6 \times 3=360$

(iii) Total numbers which are divisible by $25=40$

(iv) A number is divisible by 4 , If its last two digit is divisible by 4 .

$\therefore$ Total such numbers $=200$

Solution

(i) 4 letters are used at a time $={ }^{6} P_4=\frac{6 !}{2 !}=6 \times 5 \times 4 \times 3=360$

(ii) All letters used at a time $=6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$

(iii) All letters used but first is vowel $=2 \times 5$ ! $=2 \times 5 \times 4 \times 3 \times 2 \times 1=240$