1. ## simple factorial question

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)
How many different combinations are possible?
(I realise it's somewhere between 20 factorial and 60 factorial, but then I'm stuck.)
Expressing the answer in the form n.nn x 10 to power y would be very helpful.

Thanks
James

2. Originally Posted by james maths questioner
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

$\displaystyle {n + r - 1 \choose r}$.

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

3. ## simple factorial problem

Hi Mr F

1. Yes you're right - it's 3 of each colour - sorry.
2. I had assumed that the answer would be more than 20 factorial, because it would be 20 factorial if they ordered 20, and thay can order 60 (although the number of colours remain cosntant at 20). I'm afraid I'm a beta-brain!
3. Thank you for your answer - which is clearly billions, even in the old-fashioned UK definition of 10 to the 12th.

James