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There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to zoo. The members are to be seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How many different ways can the members travel?

Answer is given as 4c1× 7c2+ 4c2×7c1+ 4c1×7c3+ 4c2×7c2 + 4c2×7c2+ 4c1×7c3 = 658

source: question 2, http://sakshieducation.com

This is a question similar to this site, which I asked some time before and I got a good explanation from @Graham Kemp. This new question was found when I was searching for similar questions.

Answer can be

3*4C2*7C1*2C1*6C3 + 3*4C2*7C2*2*2C1*5C2 (if cars are distinct)

or

4C2*7C1*2C1*6C3 + 4C2*7C2*2C1*5C2 (if cars are identical).

So, the approach and answer given in the site referred (658) is wrong. Am I right?

Community
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Kiran
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1 Answers1

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As far as I can tell, the site is counting the number of ways to pick people to fill the first cab to arrive when it has been decided that taxi I will take 3 passengers and both the others 4 each.

This does not count the number of ways to fill all three cabs.

Graham Kemp
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