Question

Let $S=\{1,2,3,\dots ,9\}$. For $k=1,2,\dots ,5$, let $N_k$ be the number of subsets of $S$, each containing five elements out of which exactly $k$ are odd. Then $N_1+N_2+N_3+N_4+N_5=$

• A. $210$
• B. $252$
• C. $125$
• D. $126$

Answer 1

The correct option is D $126$

In the given set $1,3,5,7,9$ are odd and $2,4,6,8$ are even

Therefore there are $5$ odd and $4$ even numbers in the set

$N_1{=}^5{C}_1{\times }^4{C}_4=5$ {1 odd and 4 even}

$N_2{=}^5{C}_2{\times }^4{C}_3=40$ {2 odd and 3 even}

$N_3{=}^5{C}_3{\times }^4{C}_2=60$ {3 odd and 2 even}

$N_4{=}^5{C}_1{\times }^4{C}_4=20$ {4 odd and 1 even}

$N_5{=}^5{C}_5{\times }^4{C}_0=1$ {5 odd and 0 even}

Therefore

$N_1+N_2+N_3+N_4+N_5=5+40+60+20+1=126$