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

Question

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

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

KarishmaB · Accepted Answer

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)

JeffTargetTestPrep · Answer

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

gmatophobia · Answer

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

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

Option C

Abhishek009 · Answer

_ _ _ _

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)

ShilpiAgnihotrii · Answer

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

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

Solving this the Bunuel's way!

4 different numbers, so this can be arranged in 4! ways.
8 is repeating so divide it by 2! and 3, 5 are NOT repeating (so 1! which is esentially 1)

Ans: 4!/2! =12

Thanks to  Bunuel  for teaching this method!

kabirgandhi · Answer

Hey,

I just gave my first gmat official practice test yesterday, and got the same answer, but my score report shows that it's not correct, same happened for one more question. Is this possible? If yes, how often does it happen?

DanTheGMATMan · Answer

Set one example order and multiply that by the number of orders that could be in:

https://www.youtube.com/watch?v=QLm8ExGJku8

GMATinsight · Answer

so one possible number is 3358

But we also need to take into account all possible arrangements of these digits

Total arrangements of {3358} = 4!/2! = 12

Answer: OPtion C

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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745 with Only Self-Prep and GMAT Club

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My Rewards