Possible Combinations?

A. How many different 9 digit numbers can be formed using the digits in the number 277,728,788?

I want to use the Combination formula, since no repetition but order doesn't matter, where

9!
9!(9-9)!

but that will equal out to one... So should I do permutation?

B. How many of the above are even?

I would divide the outcome of A by two, since every other number is even.

Am I on the right track?

There are 9 digits, but 7 appears 4 times, 8 appears 3 times and 2 appears 2 times.

\(\displaystyle \frac{9!}{4!3!2!}\)

Thank you.

Hello, alex71385!

A. How many different 9-digit numbers can be formed using the digits in the number 277,728,788 ?

I want to use the Combination formula, since no repetition but order doesn't matter. . Really?

Click to expand...

\(\displaystyle \text{galactus gave you the answer: }\;\frac{9!}{4!\,3!\,2!} \:=\:1260\text{ possible 9-digit numbers.}\)

B. How many of the above are even?

I would divide the outcome of A by two, since every other number is even. . No!

Click to expand...

\(\displaystyle \text{How many of the 1260 possible numbers are }odd\,?\)

\(\displaystyle \text{To be odd, the number must end with an odd digit.}\)
. . \(\displaystyle \text{The only odd digit is }7.\)

\(\displaystyle \text{So we have a number of this form: }\;\_\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,7\)

\(\displaystyle \text{The other eight digits }\{7,7,7,8,8,8,2,2\}\text{ can be arranged in }\,\frac{8!}{3!\,3!\,2!} = 560\text{ ways.}\)
. . \(\displaystyle \text{Hence, there are 560 odd numbers.}\)

\(\displaystyle \text{Therefore, there are: }\:1260 - 560 \:=\:700\text{ even numbers.}\)