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Permutation and Combination
A father with 8 children takes 3 at a time to the zological garden, as often as he can,without taking the same 3 children more than once.how often he will go and how ofen will each child go? a)56,35 b)92,42 c)56,21 d)56,42
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- ANSWER- 21
solution:
possible no of groups (each containing 3 children) = 8C3= 56
father goes with every group hence he will go 56 times.
total children in 56 trips =56*3
as there are 8 different children, no of trips of each child=(56*3)/8=21 - 12 years agoHelpfull: Yes(40) No(1)
- ans :- c)56,21
no of possible groups= 8c3=56
hence father will go 56 times
total children in 56 groups= 56*3
there are 8 children hence no of trips for each child=(56*3)/8=21 - 12 years agoHelpfull: Yes(6) No(0)
- c)56,21
A father with 8 children takes 3 at a time to the zological garden, as often as he can,without taking the same 3 children more than once.
Number of possible groups = 8C3 = 8*7*6/6 = 56
Each child will have equal chance to go to Zoo.
So number of time , each child can go to zoo = 56*3/8 = 21
- 12 years agoHelpfull: Yes(2) No(0)
- no of visits=8C3=8!/(3!5!)=56
visit of each child=7C2=7!/(2!5!)=21 - 12 years agoHelpfull: Yes(2) No(2)
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