Question

(JEE(Advanced)-2017, 3]
14. Let S = [1,2,3......,9). For k = 1.2. ....., 5, let N, be the number of subsets of S, each containing
five elements out of which exactly k are odd. Then N. + N + N + N + Ns =
LIEE(Advanced)-2017, 3(-1)​

Answer 1

Correct Question:-

Let $S=\{1,\ 2,\ 3,\,\dots,\ 9\}.$ For $k=1,\ 2,\,\dots,\ 5,$ let $N_k$ be the no. of subsets of $S,$ each containing 5 elements out of which $k$ are odd. Then find $\displaystyle\sum N_k.$

Answer:-

$\Large\boxed{\quad126\quad}$

Solution:-

In $S,$ no. of odd elements is 5 and that of even elements is 4.

No. of subsets of $S,$ each containing 5 elements out of which 1 is odd, is,

$\longrightarrow N_1=\ ^5C_1\times\ ^4C_4$

$\longrightarrow N_1=5\times1$

$\longrightarrow N_1=5$

No. of subsets of $S,$ each containing 5 elements out of which 2 are odd, is,

$\longrightarrow N_2=\ ^5C_2\times\ ^4C_3$

$\longrightarrow N_2=10\times4$

$\longrightarrow N_2=40$

No. of subsets of $S,$ each containing 5 elements out of which 3 are odd, is,

$\longrightarrow N_3=\ ^5C_3\times\ ^4C_2$

$\longrightarrow N_3=10\times6$

$\longrightarrow N_3=60$

No. of subsets of $S,$ each containing 5 elements out of which 4 are odd, is,

$\longrightarrow N_4=\ ^5C_4\times\ ^4C_1$

$\longrightarrow N_4=5\times4$

$\longrightarrow N_4=20$

No. of subsets of $S,$ each containing 5 elements out of which 5 are odd, is,

$\longrightarrow N_5=\ ^5C_5\times\ ^4C_0$

$\longrightarrow N_5=1\times1$

$\longrightarrow N_5=1$

Hence,

$\displaystyle\longrightarrow\sum N_k=N_1+N_2+N_3+N_4+N_5$

$\displaystyle\longrightarrow \sum N_k=5+40+60+20+1$

$\displaystyle\longrightarrow \underline{\underline{\sum N_k=126}}$