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There are few boxes labelled with $1,2,4,5,7,8,9$ respectively. How many ways are there to choose $5$ boxes and arranges the boxes in a row...

a) if only one box labelled with even number can be selected?

b) if all the box labelled with even number can be selected?

Can you guys help me solve this question?

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  • $\begingroup$ a) and b) seem to be identical? $\endgroup$ – joriki Oct 10 '15 at 20:35
  • $\begingroup$ 'a' for one box 'b' for all box $\endgroup$ – Sakthi Seelen Oct 10 '15 at 20:42
  • $\begingroup$ So b) is simply without restrictions? If so, I think it would be clearer to state it that way. $\endgroup$ – joriki Oct 10 '15 at 20:46
  • $\begingroup$ The best help you can get is really to be forced to think about the problem to the point where you develop your own thoughts. $\endgroup$ – pjs36 Oct 10 '15 at 21:09
  • $\begingroup$ When you pose a question here, it is expected that you include your own attempt to solve the problem and indicate where you are stuck so that you receive responses that address the specific difficulties you are encountering. $\endgroup$ – N. F. Taussig Oct 11 '15 at 0:19
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There are 3 even numbered boxes and 4 odd numbered boxes.

a) If only one box labelled with even number can be selected, then the number of ways of choosing the boxes is (3C1) * (4C4) = 3. The number of ways to arrange these is 5! since the 5 boxes can be arranged in 5P5 ways. Hence, required answer shall be 3*5! = 360.

b) If all the even numbered boxes can be selected, then you have cases where you select 1 even 4 odd, 2 even 3 odd and 3 even 2 odd. These can be chosen in (3C1*4C4) + (3C2*4C3)+(3C3*4C2) ways, which equals 21. Again arranging them is 5! because for each case you have 5P5 ways of arranging them. Hence, the answer here will be 21*5! = 2520.

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  • $\begingroup$ Thanks for the solution..now i can understand $\endgroup$ – Sakthi Seelen Oct 11 '15 at 7:24

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