Hi, can anyone help me with this question?:

A 27 digit number consists of nine 0's, nine 1's, and nine 2's. The first digit cannot be 0. How many different numbers are possible? (leave answer in terms of nCr)

Thanks in advance!

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- Sep 23rd 2009, 10:45 AMFarside18Combination problem with digits
Hi, can anyone help me with this question?:

A 27 digit number consists of nine 0's, nine 1's, and nine 2's. The first digit cannot be 0. How many different numbers are possible? (leave answer in terms of nCr)

Thanks in advance! - Sep 23rd 2009, 11:09 AMGrandad
You could search for 'Permutations with repeated objects' Here's a typical site.

Since there are 2 choices for the first digit, and the remaining 26 digits have 9, 9 and 8 repetitions, the answer is $\displaystyle \frac{2\times26!}{9!9!8!}$.

Grandad

PS

Since the question says to leave the answer in terms of $\displaystyle {^nC_r}$, the alternative method is as follows:

There are 2 ways of choosing the first digit; then $\displaystyle {^{26}C_9}$ ways of choosing where the zeros will be positioned; then $\displaystyle {^{17}C_9}$ ways of choosing the positions occupied by the digit that didn't go into first place; the remaining 8 spaces are then filled automatically by the remaining 8 identical digits.

Total = $\displaystyle 2\times{^{26}C_9}\times{^{17}C_9}$, which you'll find comes to $\displaystyle \frac{2\times26!}{9!9!8!}$, as before.