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shrouded1
It's a tricky one, but if you look at the solution above, it's almost like saying c(n,r) will always be an integer.

Just that in this case we are talking about a different kind of arrangement with a different formula.

Posted from my mobile device

indeed very tricky, i was able to rule out (1) and guessed B

Can you explain a bit more on what you mean by "it's almost like saying c(n,r) will always be an", i'm not seeing this thru a combination? Thanks.
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Imagine I ask you the question :
What is the number of ways you can arrange A balls, each of different color in a row ?
The answer would be \(A!\)

Now I modify that question :
What is the number of ways you can arrange A balls, of which B are are blue, C are red, D are green and the rest are of different but unique colors ?
The answer would now be \(\frac{A!}{B!C!D!}\)

Now I modify it further :
What is the number of ways you can arrange A balls, of which there are x subsets each consisting B balls each such that each subset consists of balls of a different shade of blue, and all other balls not included in these subsets are of unique colors ?
The answer would now be \(\frac{A!}{B!^x}\). Notice that A has to be greater than or equal to xB

Finally I modify it a bit more :
What is the number of ways you can arrange A balls, of which there are x subsets each consisting B balls each, y subsets of C balls each, z subsets of D balls each, such that each subset consists of a unique color of balls. And the rest of the set of balls are distinct from all other balls ?
The answer would now be \(\frac{A!}{B!^xC!^yD!^z}\). Notice that A has to be greater than or equal to xB+yC+zD



Essentially what I am getting at is that the expression shown above is the answer to a combinatorial problem. And since the answer to a combinatorial problem is a "number of ways", such an expression always has to be an integer.
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Nice, thanks for explaining, now i get the picture.
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