Let S\ 1,2,3, ... ,9\. For (k = 1,2, ... ,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 =

(d): 1^ st solution : We have N_k= ^5 C_k * ^4 C_5-k \ N_1+N_2+N_3+N_4+N_5 = ^5 C_1 * ^4 C_4+ ^5 C_2 * ^4 C_3 \ + ^5 C_3 * ^4 C_2+ ^5 C_4 * ^4 C_1+ ^5 C_5 * ^4 C_0 \ =5+40+60+20+1=126 2^ nd solution : There are only 4 even numbers in S. Thus any subset of S will have at least one odd number. So, N_1+N_2+N_3+N_4+N_5 = ^9 C_5= ^9 C_4 =(9.8 * 7.6 / 24)=126

Let (S{ 1,2,3,ldots,9}). For (k = 1,2,ldots,5), let (N{... | Permutations and Combinations (Mathematics)

(d): $1^{\text {st }}$ solution : We have $\\N_k={ }^5 C_k \cdot{ }^4 C_{5-k}$ $\\ \begin{aligned} & N_1+N_2+N_3+N_4+N_5 \\&={ }^5 C_1 \cdot{ }^4 C_4+{ }^5 C_2 \cdot{ }^4 C_3 \\ & \quad+{ }^5 C_3 \cdot{ }^4 C_2+{ }^5 C_4 \cdot{ }^4 C_1+{ }^5 C_5 \cdot{ }^4 C_0 \\ & =5+40+60+20+1=126 \end{aligned}\\$ $2^{\text {nd }}$ solution : There are only 4 even numbers in $S.\\$ Thus any subset of $S$ will have at least one odd number. $\\$ So, $N_1+N_2+N_3+N_4+N_5 \\={ }^9 C_5={ }^9 C_4 \\=\frac{9.8 \cdot 7.6}{24}=126$

Explanation

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

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