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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(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

Similar Questions

It is required to seat $5$ men and $4$ women in a row so that the women occupy the even places. How many such arrangements are possible?

If $^8C_r = ^8C_{r+2}$,then the value of $^rC_2$ is

How many numbers less than $1000$ can be made from the digits $1, 2, 3, 4, 5, 6$ (repetition is not allowed)?

If $^nC_r$ denotes the number of combinations of $n$ things taken $r$ at a time,then the expression $^nC_{r+1} + ^nC_{r-1} + 2 \times ^nC_r$ equals

Assuming that no two consecutive digits are same,the number of $n$-digit numbers is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module