Question

If the number of 5 elements subset of the set A\left\\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},.....................,{{a}_{20}} \right\\} of 20 distinct elements is k times the number of 5 elements subsets containing ${{a}_{4}}$, then k is
A. 5
B. $\dfrac{20}{7}$
C. 4
D. $\dfrac{10}{3}$

Answer 1

- Hint: In general the number of ways to select r thing n number of things is $^{n}{{C}_{r}}$.
We can define $^{n}{{C}_{r}}$as below:
$^{n}{{C}_{r}}=\dfrac{n!}{r!\times (n-r)!}$

Complete step-by-step solution -
Given set is A\left\\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},.....................,{{a}_{20}} \right\\}
In this number of elements = 20
We need to choose a subset of 5 elements from a given set.
So number of ways to select 5 elements subset from 20 element set A is $^{20}{{C}_{5}}$
We can find value of $^{20}{{C}_{5}}$ as below:
${{\Rightarrow }^{20}}{{C}_{5}}=\dfrac{20!}{5!\times (20-5)!}$ \left\\{ {{\because }^{n}}{{C}_{r}}=\dfrac{n!}{r!\times (n-r)!} \right\\}
${{\Rightarrow }^{20}}{{C}_{5}}=\dfrac{20!}{5!\times 15!}$
Number of ways to select 5 subset containing ${{a}_{4}}$ is $^{19}{{C}_{4}}$.
We can find value of $^{19}{{C}_{4}}$ as below:
${{\Rightarrow }^{19}}{{C}_{4}}=\dfrac{19!}{4!\times (19-4)!}$ \left\\{ {{\because }^{n}}{{C}_{r}}=\dfrac{n!}{r!\times (n-r)!} \right\\}
${{\Rightarrow }^{19}}{{C}_{4}}=\dfrac{19!}{4!\times 15!}$
As the given number of subset of 5 elements is k times the number of 5 subset containing ${{a}_{4}}$. Hence we can write
$\Rightarrow \dfrac{20!}{5!\times 15!}=k\dfrac{19!}{4!\times 15!}$
$\Rightarrow \dfrac{20\times 19!}{5\times 4!}=k\dfrac{19!}{4!}$
$\Rightarrow \dfrac{20}{5}=k$
$\Rightarrow k=4$

Hence the required value of k is 4.

Note: In this when we select subset of 5 elements containing ${{a}_{4}}$, in this case we already selected 1 element i.e ${{a}_{4}}$, so from this total number of elements will be 19 and we need to select only 4 element of subset. We need to remember this point.
In general, the factorial of n can be defined as a product of all integers from n to 1. We can write it as
$n!=n(n-1)(n-2)(n-3)................................3.2.1$
It is defined only for positive integers.