Math, asked by RDalal, 9 months ago

(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)​

Answers

Answered by shadowsabers03
6

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}}

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