First man has 5 ways. Now the next man can stand behind the first man(and not before) or he can stand in the remaining 4 queues so he also have 5 ways. similarly all man will have 5 choices so answer should be 5^12 ways. Isn’t it? please reply.Thanks.

“First” and “Second” are just a manner of speaking. We don’t have any idea in what order the people have arrived, we just see their final position in the queue. So if A is in the first queue, and B also turns out to be there, then B could have arrived before or after A and hence in total B has 6 ways (2 in that queue and 4 in the others). This logic holds whenever we are distributing distinct objects in distinct groups, such that order within the group also matters.

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In the last question, it should be 5^12 ways.

First man has 5 ways. Now the next man can stand behind the first man(and not before) or he can stand in the remaining 4 queues so he also have 5 ways. similarly all man will have 5 choices so answer should be 5^12 ways. Isn’t it? please reply.Thanks.

“First” and “Second” are just a manner of speaking. We don’t have any idea in what order the people have arrived, we just see their final position in the queue. So if A is in the first queue, and B also turns out to be there, then B could have arrived before or after A and hence in total B has 6 ways (2 in that queue and 4 in the others). This logic holds whenever we are distributing distinct objects in distinct groups, such that order within the group also matters.

regards

J