Permutations combinatorics problem?

  • Thread starter SMA_01
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Homework Statement



A telephone extension has four digits, how many different extensions are there with no repeated digits, if the first digit cannot be zero?

Homework Equations



P(n,r)=n!/(n-r)!


The Attempt at a Solution



For the first digit, there are 9 possibilities (because no zero)
For the last 3 digits I used P(9,3)=9!/6!=9x8x7

So, in the end my result was: 9x9x8x7= 4,536 different extensions...

I'm just wondering if I was correct?
 

Answers and Replies

  • #2
Dick
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Homework Statement



A telephone extension has four digits, how many different extensions are there with no repeated digits, if the first digit cannot be zero?

Homework Equations



P(n,r)=n!/(n-r)!


The Attempt at a Solution



For the first digit, there are 9 possibilities (because no zero)
For the last 3 digits I used P(9,3)=9!/6!=9x8x7

So, in the end my result was: 9x9x8x7= 4,536 different extensions...

I'm just wondering if I was correct?

Seems fine to me.
 

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