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When arranging people (A, B, C, D, E) in seats (1,2,3,4,5), why isn't it same to count with reference to seats eg. (number of people can sit in seat number 1) * (number of people can sit in seat number 2) *.. * (number of people can sit in seat number 5)

in place of setting people as reference reference? eg.

(number of seats A can take) * (number of seats B can take) *.. * (number of seats E can take)?

Example. In how many ways can 5 girls (A, B, C, D, E) can sit on five chairs (1, 2, 3, 4, 5) if girl A does not want to sit on chair number 1 or chair number 2 and girl B wants to sit on 3rd or 4rth chair?

considering seats 3*2*3*2*1 (for seat number one to five respectively)

but when considering people

2*2*3*2*1 (for girls A, B, C, D, E)

Another example,

How to put 3 balls in 5 urns (where an urn can have any number of balls)?

I can put each ball in 5 urns: $$5^{3}$$ ways or I can select $5\choose 3$ urns and put 3! on the selected ones.

EDIT: replaced $3^{5}$ to $5^{3}$

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  • $\begingroup$ For the second case, taking $\binom{5}{3}3!$ assumes that one urn can contain at most ball. You should consider that each ball has $5$ choices of urns to be put in, so you should have $5 \times 5 \times 5 =5^3$ $\endgroup$
    – Nicholas
    Nov 21, 2015 at 17:50
  • $\begingroup$ What actually prevents me to think backwards? Also doesn't selecting one ball for the first urn reduce the number of balls that can be put on second urn? $\endgroup$
    – kuity kita
    Nov 21, 2015 at 17:55
  • $\begingroup$ The reason you can't take $3^5$ is because at any one time, one ball can only go into one box. This is why you have to think in terms of the balls not the urns. The first ball can go into any of the boxes. The second ball can go into any of the 5 urns, and so on. You can draw a tree diagram if that helps you to visualise the problem. $\endgroup$
    – Nicholas
    Nov 21, 2015 at 18:00
  • $\begingroup$ Does that mean, while counting one has to think in terms of single events? Why can't I say one urn can be filled with one ball at a time? $\endgroup$
    – kuity kita
    Nov 21, 2015 at 18:14
  • $\begingroup$ Exactly, after which, to find the total number of ways you either multiply them together or add them together depending on the case. Although, there are some problems which can't be done this way. $\endgroup$
    – Nicholas
    Nov 21, 2015 at 18:18

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