(JEE2014) Let $n \geqslant 2$ be an integer. Take $n$ distinct points on a circle and join each pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then find the value of $n$.

Solution: These are $n$ line segments joining adjacent points. So the number of blue line segments $=n$.

If we join all the points, the total number of line segments $^n{c_2}=\frac{n(n-1)}{2}$

So the number of line segment which are coloured red is : $\frac{n(n-1)}{2}-n$

Now $\frac{n(n-1)}{2}-n=n$ $$\Rightarrow n(n-1)=4 n \Rightarrow n^2=5 n \text {. So } n=5 \text {. }$$

JEE (2014) Let $n_1

Solution: $1,2,3,4,10; \quad 1,2,3,5,9 ;\quad 1,2,3,6,8$ $$1,2,4,5,8 ; \quad 1,2,4,6,7 ;\quad 1,3,4,5,7;$$ $2,3,4,5,6$.
Total number of ways $=7$.

Let $(x, y, z)$ be integers satisfying system of homogeneous equations

$3 x-y-z=0$

$-3 x+z=0$

$-3x+2 y+z=0$

Find the number of such points for which $x^2+y^2+z^2 \leq 100.$

Solution:
Here $y=0, \quad z=3 x$

$x^2+z^2 \leqslant 100$

Only possible three truples $(x, y, z)$ as

$(0,0,0),(1,0,3),(-1,0,-3),$

$(2,0,6),(-2,0,-6),(3,0,9),(-3,0,-9)$

So the total number of integer solutions 7 .

(JEE 2014). Three boys and two girls stand in a queue. In how many ways they can be arranged so that the number of boys ahead of every girl is at least one more than the number of girls ahead of her ?

Solution: Let boys be $B_1, B_2, B_3$ and girls be $G_1, G_2$. We can have following placements.

(i)$G_1 G_2 B_1 B_2 B_3$ (We can permute 2 girls in 2 ! ways & 3 boys in 3 ! ways,) so total number of such arrangements is $2 ! \times 3 !=12$

(ii) $G_1 B_1 G_2 B_2 B_3$ Again, boys can be permuted in 3! ways & girls can be permuted in $2 !$ ways

So total number of arrangements is $3 ! \times 2 !=12$

(iii) $G_1 B_1 B_2 G_2 B_3 \rightarrow 3 ! \times 2 !=12$ such arrangements.

(iv) $B_1 G_1 G_2 B_2 B_3 \rightarrow 3 ! \times 2 !=12$ such arrangements.

(v) $B_1 G_1 B_2 G_2 B_3 \longrightarrow 3 ! \times 2 !=12$ such arrangements.

such So total number of such arrangements $=12 \times 5=60$

Consider word ’ ENDEANOEL'.

(i) How many words can be formed by considering all letters of this word?

$9 \rightarrow \text { letters, } E \rightarrow 3, N \rightarrow 2$

$A, D, O, L \rightarrow 1$

$\frac{9 !}{3 ! \times 2 !}=30240$

(ii) In how many words the segment ‘ENDEA’ occurs?

Here the number of objects is 5 So the number of arrangements is $5 !=120$.

(iii) The number of words starting and ending with ’ $E$ '

So now we have 7 letters remaining in which N occurs twice:

$\frac{7!}{2!}=2520$

(iv) In how many words letters $A, E, O$ occur only in odd positions?

There are $3 E^{\prime}$ s. Total odd positions are $1,3,5,7,9$. So in 5 places, $A, O$ and $3 E^{\prime} s$ can be placed in $\frac{5 !}{3 !}$ ways.

In remaining 4 positions $L, D \ \text{and} \ 2 \mathrm{~N}$ ’s can be placed in $\frac{4 !}{2 !}$ ways.

So total number of such words $=\frac{5 !}{3 !} \times \frac{4 !}{2 !}=240$

(v) The number of permutations in which none of letters $D, L, N$ occur in the last five positions.

$D, L,$&$2 \mathrm{~N}$ ’s must appear in first four places $\rightarrow \frac{4 !}{2 !}$

Remaining $A, O, 3 E^{\prime}$ s can be placed in last 5 positions in $\frac{5 !}{3 !}$ ways.

$\frac{4 !}{2 !} \times \frac{5 !}{3 !}=240$

(JEE 2015) In how many ways 5 boys and 5 girls can be placed is a queue so that

(i) all girls stand consecutively ?

(ii) exactly 4 girls stand consecutively ?

Solution:

(i) If all the girls stand consecutively, they can be treated as one unit. So total number of permutations of 5 boys +1 unit is 6 !.

However, the 5 girls can be permuted in 5 ! ways. So the total number of such arrangements is $6 ! \times 5 !=86400$.

(ii) 4 girls who stand consecutively, can be treated as one unit. So 5 boys and these 4 girls can stand in 6! ways.

Further these 4 girls can be permuted in 4 ! ways. The girl who stands separately can be choosen from 5 girls in ${ }^5 C_1$ ways.

The last girl can be placed in 5 ways (as it can be placed adjacent to previous 4 girls). So now by multiplication principle, the total number of arrangements

$=6 ! \times 4 ! \times{ }^5 C_9 \times 5=432000$

Lisa chooses 13 cards from a deck of 52 cards. In how many ways she can choose so that she gets 2 kings & 2 queens.

Solution:

${ }^4 c_2 \times{ }^4 c_2 \times{ }^{44} c_9$

All permutations of the word ’ $L A S T ‘$ are arranged according to english dictionary ordering. what is the positionss of the word ’ $S A L T$ ’ in this ordering?

Solution: The number of words starting with

$$\text { A } \square \square \square \rightarrow 3 !=6$$

The number of words starting with

$$\text { L } \square \square \square \rightarrow 3 !=6$$

Next one will be $L A S T^{\prime}$ $\rightarrow 13^{\text {th }}$ positions