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sritamasia
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Status:Applying to ISB EEO, HEC Paris MiM, ESSEC MIM, SPJIMR PGDM
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Concentration: Marketing, Strategy
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dave13
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So you can simple say it is 25P15 instead of 25C15 and then multiplying by 15!

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Dhwanii
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uttamkgulshan so like I am not sure how to decide if 25P15, will also take care of arranging 10 empty squares in 10! ways and 15 numbers in 15! ways
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So the empty squares need not be arranged in 10! Ways because they will remain indistinguishable. The only thing we need is permutation of 15 numbers in 25 boxes. That's it

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Dhwanii
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uttamkgulshan thanks for your help, I agree 10 empty squares are identical but whether we leave first 2 rows empty or last 2 columns etc. are different cases na hence I was saying 10! ways
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Just think that there are 25 boxes and 15 different balls or numbers. They will be arranged as per permutation. Now, you put theses boxes in the form of square, or rectangle or hexagon, doesn't matter. Last column, row etc every possibility will be covered in these permutations

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ZIX
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If there is still confusion, following abstraction would help:

First, we need to choose which 15 squares (out of 25) will contain numbers.
This is: 25C15

Once we've chosen which 15 squares get numbers, we need to assign the numbers 1, 2, 3, ..., 15 to these squares.
Since the numbers are all distinct, this is a permutation of 15 distinct objects in 15 positions: 15!

So total ways = 25C15 × 15!
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