If the number of 5-element subsets of the set | vedclass

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(C) The set $A = \{a_1, a_2, \dots, a_{20}\}$ contains $20$ distinct elements.The total number of $5$-element subsets is given by the combination form

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The set $A = \{a_1, a_2, \dots, a_{20}\}$ contains $20$ distinct elements.The total number of $5$-element subsets is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.Thus,the total number of $5$-element subsets is $\binom{20}{5} = \frac{20!}{5!15!}$.Now,we find the number of $5$-element subsets that contain $a_4$. If $a_4$ is already included,we need to choose $4$ more elements from the remaining $19$ elements $(20 - 1 = 19)$.So,the number of $5$-element subsets containing $a_4$ is $\binom{19}{4} = \frac{19!}{4!15!}$.According to the problem,the total number of subsets is $k$ times the number of subsets containing $a_4$:$\binom{20}{5} = k 	imes \binom{19}{4}$$\frac{20!}{5!15!} = k 	imes \frac{19!}{4!15!}$$\frac{20}{5} 	imes \frac{19!}{4!15!} = k 	imes \frac{19!}{4!15!}$$4 = k$Therefore,the value of $k$ is $4$.

If the number of $5$-element subsets of the set $A = \{a_1, a_2, \dots, a_{20}\}$ of $20$ distinct elements is $k$ times the number of $5$-element subsets containing $a_4$,then $k$ is

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