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January 28, 2015

January 28, 2015

Posted by **Jane** on Sunday, September 9, 2007 at 11:27pm.

- precalculus -
**drwls**, Monday, September 10, 2007 at 12:16amThat depends upon the maximum number of toppings per slice. With N varieties of topping, and the total number of chosen toppings running from 0 to N, the possible number of toppings is

1 + N + N*(N-1)/2! + N(N-1)(N-2)/3! + .. + N + 1

The series continues for each N until you end up with last terms of N and 1. Try various value of N unti you exceed 4000 for the series sum.

For N = 10, the number of possibilities is

1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1024. This happens to equal 2^10. Although I have not proived it, the number of possible topping combinations with N toppings appears to ne 2^N. With 12 toppings, 4096 combinations are possible.

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