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wisterville · #4 August 21st, 2008, 02:04 AM

Hello,

It is not so complicated once you get it...

For using the calculator, the formulae
${}_nH_r={}_{n+r-1}C_r$ and ${}_nC_r=\frac{n!}{(n-r)!r!}$ and $r!=r\cdot(r-1)\cdots 3\cdot2\cdot 1$ should do.

If the 20 dices were replaced by 20 people, you can distinguish between them so it becomes a permutation. In this case, you choose 20 (maybe same)elements out of 3 ones (The order of choice is not arbitrary).
Each person can choose from 3 alternatives, so ${}_3\Pi_{20}=3^{20}$ .

If 10 people have only 2 alternatives, the total would be
${}_2\Pi_{10}\cdot{}_3\Pi_{10}=2^{10}\cdot 3^{10}$ .

Bye.