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Q.

A father with 8 children takes 3 children at a time to the zoological garden, as often as he can without taking the same 3 children together more than once. Then:

 A.

number of times he will go to zoological garden is 56.

 B.

number of times each child will go to the zoological garden is 21.

 C.

number of times a particular child will not go to the zoological garden is 35

 D.

All of the above.

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Solution:
Option(D) is correct

The number of times the father would go to the zoological garden = Number of ways of selection of 3 children taken at a time is

\(={^8C_3}=56\)

Number of times a child will go to the zoological garden = Number of times he is accompanied by two other

\(=1\times {^7C_2}=21\)

Number of times a child will not go to the zoological garden:

$=56−21= 35$

Edit: For an alternative approach, check comment by Sravan Reddy.


(3) Comment(s)


Aqsa Hameed
 ()

Is combination included in Algebra?



Sravan Reddy
 ()

Alternate way to approach 2nd option in case ${^7C_2}$ does not strike (may take similar or more time):

=> Total number of times he will go to zoo - 56.

=> Total children taken including repetitions - 56*3

=> Due to symmetry, each child will go $\dfrac{56*3}{8} = 21 \text{ times}$



PRATYUSH ANAND
 ()

This is really good question. But as it's topic is Algebra never thought of computation.