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Permutation and Combination -Letter Arrangements

Problem Solving — algebra and arithmetic
Expert replies
by s91arvindh » Fri May 02, 2014 5:49 am
List A: C, M, G, L, S, U,o
List B: X, Y, Z, P, T

How many different 4-letter arrangements can be made by selecting 2 letters each from list A and list B above?
a) 4500 b) 3060 c) 2930 d) 5040

Answer is D

Derivation :

Selection of 2 letters from list A 7*6 ways

Selection of 2 letters from list B 5*4 ways

Total ways will be (7*6)*(5*4)
Stuck here :(
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Source: — Quantitative Reasoning |

by Brent@GMATPrepNow » Fri May 02, 2014 6:14 am
s91arvindh wrote: List A: C, M, G, L, S, U, O
List B: X, Y, Z, P, T

How many different 4-letter arrangements can be made by selecting 2 letters each from list A and list B above?
a) 4500 b) 3060 c) 2930 d) 5040
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.


Take the task of selecting and arranging the letters, and break it into stages.

Stage 1: Select 2 letters from list A.
Since the order in which we SELECT the two letters does not matter, we can use combinations.
We can select 2 letters from 7 letters in 7C2 ways (21 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 7C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select 2 letters from list B.
We can select 2 letters from 5 letters in 5C2 ways (10 ways)

Now that we've selected the 4 letters we'll use to create our word, it's just a matter of arranging them.

Stage 3: Arrange the 4 letters.
We have a nice rule for this. We can arrange n unique objects in a row in n! ways.
So, we can complete stage 3 in 4! (24 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a 4-letter word) in (21)(10)(24) ways ([spoiler]= 5040 ways[/spoiler])

Answer: D

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Last edited by Brent@GMATPrepNow on Fri May 02, 2014 6:35 am, edited 1 time in total.
Brent Hanneson - Creator of GMATPrepNow.com
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by s91arvindh » Fri May 02, 2014 6:25 am
Hi Brent,

Thanks a lot for clarification :)
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