Out of 11 members of a family 4 are males and rest females. The family hired three taxis on rent to see a museum. The members have to sit in the cars in such a way that not more than 4 members are in any one car and each car has at least one male member. In how many different ways the members can travel?
Correct Answer: Description for Correct answer:
Total members in the family = 11
Number of males = 4
Number of females = 7
Number of cars hired = 3
No car can have more than 4 members. So the possible combinations can be such that one car has 3 members and the other two cars 4 each.
Further, each car must have at least one male. The total no. of males is 4. So, only one car will have 2 males and the rest one each.
Option : I II III
Total members Possible : 3 4 4
combination : 1 m 1m 2m
and 1f and 3f and 2f
or or or
(2m 2m 1m
and 1f and 2f and 3f) x 2
Total number of ways
= \( \Large \left(4_{C_{1}} \times 7_{C_{2}} + 4_{C_{1}} \times 7_{C_{3}} + 4_{C_{2}} \times 7_{C_{2}} + 4_{C_{2}} \times 7_{C_{1}}+4_{C_{2}} \times 7_{C_{2}} + 4_{C_{1}} \times 7_{C_{2}}\right) \times 2 \)
= \( \Large \left[ 4 \times 21 + 4 \times 35 + 6 \times 21 + 6 \times 7 + 6 \times 21 + 4 \times 35 \right] \times 2 \)
= 126 + 266 + 266
Total No. of ways = 126 + 266 + 266 = 658
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