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

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(C) The set $A = \{a_1, a_2, \ldots, a_{20}\}$ has $20$ distinct elements.The total number of $5$-element subsets is given by $\binom{20}{5}$.The numb

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$4$

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(C) The set $A = \{a_1, a_2, \ldots, a_{20}\}$ has $20$ distinct elements.The total number of $5$-element subsets is given by $\binom{20}{5}$.The number of $5$-element subsets containing $a_4$ is equivalent to choosing $4$ more elements from the remaining $19$ elements,which is $\binom{19}{4}$.According to the problem,$\binom{20}{5} = k 	imes \binom{19}{4}$.Using the formula $\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$,we have $\binom{20}{5} = \frac{20}{5} \binom{19}{4} = 4 \binom{19}{4}$.Comparing this with $k \binom{19}{4}$,we get $k = 4$.

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

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