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Problem:

How many different positive, five-digit integers can be formed using the digits 2, 2, 2, 9 and 9?

rohitaop  May 11, 2018
 #1
avatar
+2

If you want ONLY distinct permutations, then you can have:

5! / 3! . 2! =10 distinct permutations as follows:

{2, 2, 2, 9, 9} | {2, 2, 9, 2, 9} | {2, 2, 9, 9, 2} | {2, 9, 2, 2, 9} | {2, 9, 2, 9, 2} | {2, 9, 9, 2, 2} | {9, 2, 2, 2, 9} | {9, 2, 2, 9, 2} | {9, 2, 9, 2, 2} | {9, 9, 2, 2, 2} (total: 10)

Guest May 11, 2018
 #2
avatar+2765 
+2

_ _ _ _ _

We have the numbers 2,2,2,9,9

Possibilities with 2: 2,2,2,9,9

2,2,9,2,9, 2,2,9,9,2, 2,9,2,2,9, 2,9,2,9,2, 2,9,9,2,2

We have only four possibilties, because we have only two 9's.

Thus, the answer is \(6+4=\boxed{10}\)

smileysmiley

tertre  May 12, 2018

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