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All of the telephone extensions in a certain company are 4-digit

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MasteringGMAT
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All of the telephone extensions in a certain company are 4-digit [#permalink] New post   02 Jul 2023, 07:05
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All of the telephone extensions in a certain company are 4-digit numbers. How many different 4-digit telephone extensions can be formed from the digits 3, 5, and 8, if in each telephone extension the digits 5 and 8 are to appear once and the digit 3 is to appear twice?

A. 6
B. 10
C. 12
D. 16
E. 24
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C
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Re: All of the telephone extensions in a certain company are 4-digit [#permalink] New post   02 Jul 2023, 07:16
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MasteringGMAT wrote:
All of the telephone extensions in a certain company are 4-digit numbers. How many different 4-digit telephone extensions can be formed from the digits 3, 5, and 8, if in each telephone extension the digits 5 and 8 are to appear once and the digit 3 is to appear twice?

A. 6
B. 10
C. 12
D. 16
E. 24


The number of codes formed are various arrangements of 3, 3, 5, and 8

\(= \frac{4! }{ 2! } = 12\)

Option C
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Re: All of the telephone extensions in a certain company are 4-digit [#permalink] New post   04 Jul 2023, 04:30
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MasteringGMAT wrote:
All of the telephone extensions in a certain company are 4-digit numbers. How many different 4-digit telephone extensions can be formed from the digits 3, 5, and 8, if in each telephone extension the digits 5 and 8 are to appear once and the digit 3 is to appear twice?

A. 6
B. 10
C. 12
D. 16
E. 24


The possible extensions are permutations of the digits 3, 3, 5, and 8. Since there are identical items in the list, we use the permutation with repetition formula:

4!/2! = 4 × 3 = 12

Answer: C
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Re: All of the telephone extensions in a certain company are 4-digit [#permalink] New post   04 Jul 2023, 07:51
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MasteringGMAT wrote:
All of the telephone extensions in a certain company are 4-digit numbers. How many different 4-digit telephone extensions can be formed from the digits 3, 5, and 8, if in each telephone extension the digits 5 and 8 are to appear once and the digit 3 is to appear twice?

A. 6
B. 10
C. 12
D. 16
E. 24

_ _ _ _

Fix 5 in 4! ways
Fix 8 in 3! ways after placing 5

Since 3 is appearing twice and is being repeated, the possible no of ways is 4!*3!*2/2 = 12 , Answer must be (C)
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Re: All of the telephone extensions in a certain company are 4-digit [#permalink] New post   08 Apr 2024, 08:10
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MasteringGMAT wrote:
All of the telephone extensions in a certain company are 4-digit numbers. How many different 4-digit telephone extensions can be formed from the digits 3, 5, and 8, if in each telephone extension the digits 5 and 8 are to appear once and the digit 3 is to appear twice?

A. 6
B. 10
C. 12
D. 16
E. 24

This is a question on arrangements with some objects identical. 
When we have to arrange n objects and r of them are identical, we can arrange them in n!/r! ways. 

Here we have to use 3 twice and 5 and 8 once each. So codes will be like 3358, 3538 etc. How many such arrangements are there?
\(\frac{4!}{2!} = 12\)

Answer (C)
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Re: All of the telephone extensions in a certain company are 4-digit [#permalink]
08 Apr 2024, 08:10
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