A father with 8 children takes them 3 at a ti | vedclass

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(B) The total number of ways to select $3$ children out of $8$ is given by the combination formula ${^n}{C_r} = \frac{n!}{r!(n-r)!}$.Here,$n = 8$ and

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$21$

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(B) The total number of ways to select $3$ children out of $8$ is given by the combination formula ${^n}{C_r} = \frac{n!}{r!(n-r)!}$.Here,$n = 8$ and $r = 3$,so the total number of trips is ${^8}{C_3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.Each trip involves $3$ children. Therefore,the total number of 'child-trips' is $56 	imes 3 = 168$.Since there are $8$ children and each child goes an equal number of times,the number of times each child goes is $\frac{168}{8} = 21$.Alternatively,for a specific child to be included in a group of $3$,we must choose $2$ more children from the remaining $7$ children. This is given by ${^7}{C_2} = \frac{7 	imes 6}{2 	imes 1} = 21$.

$A$ father with $8$ children takes them $3$ at a time to the Zoological gardens,as often as he can without taking the same $3$ children together more than once. The number of times each child will go to the garden is

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