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

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(C) The set $A = \{a_1, a_2, \ldots, a_{20}\}$ contains $20$ distinct elements.The total number of $5$-element subsets is given by ${}^{20}C_5$.The nu

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$4$

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(C) The set $A = \{a_1, a_2, \ldots, a_{20}\}$ contains $20$ distinct elements.The total number of $5$-element subsets is given by ${}^{20}C_5$.The number of $5$-element subsets containing a specific element $a_4$ is equivalent to choosing $4$ more elements from the remaining $19$ elements,which is ${}^{19}C_4$.According to the problem,${}^{20}C_5 = k 	imes {}^{19}C_4$.Using the formula ${}^{n}C_r = \frac{n}{r} 	imes {}^{n-1}C_{r-1}$,we have:${}^{20}C_5 = \frac{20}{5} 	imes {}^{19}C_4$.Comparing this with ${}^{20}C_5 = k 	imes {}^{19}C_4$,we get $k = \frac{20}{5} = 4$.

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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