Question

Let S = {1, 2, 3, …, 9}. Then 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 what is the value of \[{{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}+{{N}_{5}}\]?
(a) 210
(b) 252
(c) 125
(d) 126

Answer 1

Hint: To solve this problem we will divide this problem into 5 cases for the five values of k. Like in the first case i.e. when k = 1, in that case we have to find the number of five elements chosen from
S = {1, 2, 3, …, 9} out of which exactly 1 is odd so in this case we have to select 4 even elements from S and 1 element, even elements are {2, 4, 6, 8} and odd elements are {1, 3, 5, 7, 9} hence we can select 4 elements from 4 even numbers in $^{4}{{C}_{4}}$ ways and 1 element from 5 odd numbers in $^{5}{{C}_{1}}$ ways. Similarly, we will find this for all cases and add them to get the answer.

Complete step-by-step answer:
We are given S = {1, 2, 3, …, 9} and we have to find \[{{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}+{{N}_{5}}\] where ${{N}_{k}}$ for
k = 1, 2, …,5 is the subset of S containing 5 elements out of which exactly k are odd.
So to solve this problem we will break it into 5 parts for each value of k.
So, case 1: k = 1
We have,
S = {1, 2, 3, …, 9}
Even = {2, 4, 6, 8} and odd = {1, 3, 5, 7, 9},
Now we have to make a subset out of S which contains 1 odd element and 4 even element so,
Number of ways of selecting 4 even elements from set of even elements are = $^{4}{{C}_{4}}$, and
Number of ways of selecting 4 even elements from set of even elements are = $^{5}{{C}_{1}}$
So we get total number of ways of making a subset of S which contains 5 elements from which exactly 1 is odd is i.e. ${{N}_{1}}$ = $^{4}{{C}_{4}}{{\times }^{5}}{{C}_{1}}$

Now we will proceed in same way for other cases also and find the ${{N}_{2}},{{N}_{3}},{{N}_{4}}\,and\,{{N}_{5}}$
Case 2: k = 2
We have to select 3 even elements and 2 odd elements, so we get
${{N}_{2}}{{=}^{4}}{{C}_{3}}{{\times }^{5}}{{C}_{2}}$
Case 3: k = 3
We have to select 2 even elements and 3 odd elements, so we get
${{N}_{3}}{{=}^{4}}{{C}_{2}}{{\times }^{5}}{{C}_{3}}$
Case 4: k = 4
We have to select 1 even elements and 4 odd elements, so we get
${{N}_{4}}{{=}^{4}}{{C}_{1}}{{\times }^{5}}{{C}_{4}}$
Case 5: k = 5
We have to select all 5 odd elements.
${{N}_{5}}{{=}^{4}}{{C}_{0}}{{\times }^{5}}{{C}_{5}}$
Now from all the cases, we get
\[{{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}+{{N}_{5}}{{=}^{4}}{{C}_{4}}{{\times }^{5}}{{C}_{1}}{{+}^{4}}{{C}_{3}}{{\times }^{5}}{{C}_{2}}{{+}^{4}}{{C}_{2}}{{\times }^{5}}{{C}_{3}}{{+}^{4}}{{C}_{1}}{{\times }^{5}}{{C}_{4}}{{+}^{4}}{{C}_{0}}{{\times }^{5}}{{C}_{5}}\]
Using $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we get
\[\begin{align}
  & {{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}+{{N}_{5}}=1\times 5+4\times 10+6\times 10+4\times 5+1\times 1 \\
 & {{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}+{{N}_{5}}=5+40+60+20+1 \\
 & {{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}+{{N}_{5}}=126 \\
\end{align}\]
Hence we get option (d) as the correct answer.

Note: Many students make mistakes while making the cases and select the number of odd elements directly from the set S, which is wrong so we have to select it from the set of odd elements. So you need to be careful while doing the selection of the elements. And also we used the expansion $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ in the above solution, and you should remember it for other problems also.