Answer :

3-digit numbers have to be formed using the digits 1 to 9 .

Here, the order of the digits matters.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, required number of 3-digit numbers

$ ={ }^{9} P_3=\dfrac{9 !}{(9-3) !}=\dfrac{9 !}{6 !} $

$=\dfrac{9 \times 8 \times 7 \times 6 !}{6 !}=9 \times 8 \times 7=504$

Answer :

There are four places for the digits in a 4-digit number: thousands, hundreds, tens, and units. We can fill these places one by one following these constraints:

1. Thousands place: The digit cannot be zero, because then the number would become a 3-digit number. So, there are 9 choices $(1, 2, 3, 4, 5, 6, 7, 8, 9)$ for the thousands place.

2. Hundreds place: Since no digit can be repeated, there are only $9$ digits remaining to fill the hundreds place.

3. Tens place: After filling the thousands and hundreds places, there are 8 digits remaining to fill the tens place.

4. Units place: Finally, there are $7$ digits remaining to fill the units place.

By the fundamental counting principle, the total number of unique 4-digit numbers possible is the product of the number of choices at each place:

Total numbers = Choices (thousands) × Choices (hundreds) × Choices (tens) × Choices (units) $= 9 × 9 × 8 × 7$

Therefore, there are $4536$ ($9$ x $504)$ different 4-digit numbers that can be formed with no digits repeated.

Answer :

3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits.

Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6 .

Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.

Number of ways of filling hundreds and tens place

$ ={ }^{5} P_2=\dfrac{5 !}{(5-2) !}=\dfrac{5 !}{3 !} $

$=\dfrac{5 \times 4 \times 3 !}{3 !}=20$

Thus, by multiplication principle, the required number of 3-digit numbers is

$3 \times 20=60$

Answer :

4-digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5 .

There will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

Therefore, required number of 4 digit numbers $=$

$ { }^{5} P_4=\dfrac{5 !}{(5-4) !}=\dfrac{5 !}{1 !} $

$=1 \times 2 \times 3 \times 4 \times 5=120$

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4 .

The number of ways in which units place is filled with digits is 2 .

Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits.

Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.

Number of ways of filling the remaining places

$ ={ }^{4} P_3=\dfrac{4 !}{(4-3) !}=\dfrac{4 !}{1 !} $

$=4 \times 3 \times 2 \times 1=24$

Thus, by multiplication principle, the required number of even numbers is

$24 \times 2=48$

Answer :

From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.

Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.

Thus, required number of ways $=$

$ ={ }^{8} P_2=\dfrac{8 !}{(8-2) !}=\dfrac{8 !}{6 !}=\dfrac{8 \times 7 \times 6 !}{6 !}=8 \times 7=56 $

Answer :

$ \begin{aligned} & { }^{n-1} P_3:{ }^{n} P_4=1: 9 \\ & \Rightarrow \dfrac{{ }^{n-1} P_3}{{ }^{n} P_4}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{[\dfrac{(n-1) !}{(n-1-3) !}]}{[\dfrac{n !}{(n-4) !}]}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{(n-1) !}{(n-4) !} \times \dfrac{(n-4) !}{n !}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{(n-1) !}{n \times(n-1) !}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{1}{n}=\dfrac{1}{9} \\ & \therefore n=9 \end{aligned} $

Answer:

(i)

$ \begin{aligned} & { }^{5} P_r=2^{6} P _{r-1} \\ & \Rightarrow \dfrac{5 !}{(5-r) !}=2 \times \dfrac{6 !}{(6-r+1) !} \\ & \Rightarrow \dfrac{5 !}{(5-r) !}=\dfrac{2 \times 6 !}{(7-r) !} \\ & \Rightarrow \dfrac{5 !}{(5-r) !}=\dfrac{2 \times 6 \times 5 !}{(7-r)(6-r)(5-r) !} \\ & \Rightarrow 1=\dfrac{2 \times 6}{(7-r)(6-r)} \\ & \Rightarrow(7-r)(6-r)=12 \end{aligned} $

$ \begin{aligned} & \Rightarrow 42-6 r-7 r+r^{2}=12 \\ & \Rightarrow r^{2}-13 r+30=0 \\ & \Rightarrow r^{2}-3 r-10 r+30=0 \\ & \Rightarrow r(r-3)-10(r-3)=0 \\ & \Rightarrow(r-3)(r-10)=0 \\ & \Rightarrow(r-3)=0 \text{ or }(r-10)=0 \\ & \Rightarrow r=3 \text{ or } r=10 \end{aligned} $

It is known that,

$ { }^{n} P_r=\dfrac{n !}{(n-r) !} \text{, where } 0 \leq r \leq n $

$\therefore 0 \leq r \leq 5$

Hence, $r \neq 10$

$\therefore r=3$

(ii)

$ \begin{aligned} & { }^{5} P_r={ }^{6} P _{r-1} \\ & \Rightarrow \dfrac{5 !}{(5-r) !}=\dfrac{6 !}{(6-r+1) !} \\ & \Rightarrow \dfrac{5 !}{(5-r) !}=\dfrac{6 \times 5 !}{(7-r) !} \\ & \Rightarrow \dfrac{1}{(5-r) !}=\dfrac{6}{(7-r)(6-r)(5-r) !} \\ & \Rightarrow 1=\dfrac{6}{(7-r)(6-r)} \\ & \Rightarrow(7-r)(6-r)=6 \\ & \Rightarrow 42-7 r-6 r+r^{2}-6=0 \\ & \Rightarrow r^{2}-13 r+36=0 \\ & \Rightarrow r^{2}-4 r-9 r+36=0 \\ & \Rightarrow r(r-4)-9(r-4)=0 \\ & \Rightarrow(r-4)(r-9)=0 \\ & \Rightarrow(r-4)=0 \text{ or }(r-9)=0 \\ & \Rightarrow r=4 \text{ or } r=9 \end{aligned} $

It is known that,

$ { }^{n} P_r=\dfrac{n !}{(n-r) !} \text{, where } 0 \leq r \leq n $

Hence, $r \neq 9$

$\therefore r=4$

Answer :

There are 8 different letters in the word EQUATION.

Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of 8 different objects taken 8 at a time, which is ${ }^{8} P_8=8$ !.

Thus, required number of words that can be formed $=8 !=40320$

Answer :

There are 6 different letters in the word MONDAY.

(i) Number of 4-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of 6 different objects taken 4 at a time, which is ${ }^{6} P_4$.

Thus, required number of words that can be formed using 4 letters at a time is

$ { }^{6} P_4=\dfrac{6 !}{(6-4) !}=\dfrac{6 !}{2 !}=\dfrac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}=6 \times 5 \times 4 \times 3=360 $

(ii) Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of 6 different objects taken 6 at a time, which is ${ }^{6} P_6=6$ !

Thus, required number of words that can be formed when all letters are used at a time $=6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=$ 720

(iii) In the given word, there are 2 different vowels, which have to occupy the rightmost place of the words formed. This can be done only in 2 ways.

Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. This can be done in 5 ! ways.

Thus, in this case, required number of words that can be formed is

$5 ! \times 2=120 \times 2=240$

Answer :

In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word

$ \begin{aligned} & =\dfrac{11 !}{4 ! 4 ! 2 !} \\ & =\dfrac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1} \\ & =\dfrac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1 \times 2 \times 1} \\ & =34650 \end{aligned} $

There are 4 Is in the given word. When they occur together, they are treated as a single object IIII for the time being. This single object together with the remaining 7 objects will account for 8 objects.

These 8 objects in which there are 4 Ss and 2 Ps can be arranged in $\dfrac{8 !}{4 ! 2 !}$ ways i.e., 840 ways.

Number of arrangements where all Is occur together $=840$

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four Is do not come together $=34650$-$840=33810$

Answer :

In the word PERMUTATIONS, there are $2 Ts$ and all the other letters appear only once.

(i) If $P$ and $S$ are fixed at the extreme ends ( $P$ at the left end and $S$ at the right end), then 10 letters are left.

Hence, in this case, required number of arrangements

$ =\dfrac{10 !}{2 !}=1814400 $

(ii) There are 5 vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects. These 8 objects in which there are 2 Ts can be

arranged in $\dfrac{8 !}{2 !}$ ways .

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5 ! ways.

Therefore, by multiplication principle, required number of arrangements in this case $=\dfrac{8 !}{2 !} \times 5 !=2419200$

(iii) The letters have to be arranged in such a way that there are always 4 letters between $P$ and $S$.

Therefore, in a way, the places of $P$ and $S$ are fixed. The remaining 10 letters in which there are $2 Ts$ can be arranged in $\dfrac{10 !}{2 !}$ ways

Also, the letters $P$ and $S$ can be placed such that there are 4 letters between them in $2 \times 7=14$ ways.

Therefore, by multiplication principle, required number of arrangements in this case

$ =\dfrac{10 !}{2 !} \times 14=25401600 $