Thread: simple factorial question
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Old May 2nd, 2008, 05:31 AM
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Originally Posted by james maths questioner View Post
Hi All

I have 20 different coloured bangles.
Customers buy 60, in any combination.
(e.g. 60 of same colour, 5 each of all 20 colours, etc) Mr F says: 5 each of all 20 colours is 100, not 60 .....?
How many different combinations are possible?
(I realise it's somewhere between 20 factorial and 60 factorial, but then I'm stuck.) Mr F asks: why do you think this?
Expressing the answer in the form n.nn x 10 to power y would be very helpful.

Thanks
James
This is a combinations with replacement problem:

With repetition allowed, the number of different combinations of r objects chosen from n distinguishable objects is

{n + r - 1 \choose r}.

So for your problem I get approximately 8.83829 \times 10^{18} (a number which is less than 20! by the way).
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