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(B) The total number of subsets with $8$ elements from a set of $n$ elements is given by $\binom{n}{8}$.The number of subsets with $8$ elements that c

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) The total number of subsets with $8$ elements from a set of $n$ elements is given by $\binom{n}{8}$.The number of subsets with $8$ elements that contain $a_4$ is equivalent to choosing $7$ more elements from the remaining $(n-1)$ elements,which is $\binom{n-1}{7}$.According to the problem,$\binom{n}{8} = 5 	imes \binom{n-1}{7}$.Using the identity $\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$,we have $\binom{n}{8} = \frac{n}{8} \binom{n-1}{7}$.Substituting this into the equation: $\frac{n}{8} \binom{n-1}{7} = 5 	imes \binom{n-1}{7}$.Since $\binom{n-1}{7} 
eq 0$,we can divide both sides by $\binom{n-1}{7}$ to get $\frac{n}{8} = 5$.Therefore,$n = 5 	imes 8 = 40$.

If the number of subsets with $8$ elements from the set $A=\{a_1, a_2, a_3, \ldots, a_n\}$,$n \geq 8$,is five times the number of such subsets containing $a_4$,then $n=$

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