Easy Algebra Solved QuestionAptitude Discussion

 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.

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.