Math Is Fun Forum

goed

Member

Registered: 2008-02-05

Posts: 4

Permutations and Combinations

How to find the total number of permutations of four letters selected from the word 'ARRANGEMENT'.How many of these permutations contain two R's??

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Permutations and Combinations

Case 1: No repeated letter.
This would be the number of permutations of 4 letters from the 7 distinct letters in the word, which is [sup]7[/sup]P[sub]4[/sub].

Case 2: Exactly one repeated letter.
Suppose the A is repeated. The two A’s can space themselves out in the four-letter word in 6 ways (AA**, A*A*, A**A, *AA*, *A*A, **AA). Then the other two spots are filled by 2 letters from the other 6, so the number of permutations in which the A is repeated is 6×[sup]6[/sup]P[sub]2[/sub]. This goes for the E, N and R as well. Hence the total number of permutations containing exactly one repeated letter is 4×6×[sup]6[/sup]P[sub]2[/sub].

Case 3: Two repeated letters.
Such a word can only contain the letters A, E, N, R. We choose two letters from them in [sup]4[/sup]C[sub]2[/sub] = 6 ways, and the two letters can arrange themselves in the word in 6 ways. So the total number of such words is 6×6.

Hence the total number of 4-letter permutations is sum of the permutations found in Cases 1–3.

Of these, the number of permutations containing two R’s and two different other letters is 6×[sup]6[/sup]P[sub]2[/sub]. The number of permutations containing two R’s and two A’s or two E’s or two N’s is 6×[sup]3[/sup]C[sub]2[/sub] = 6×3. Add the two to find the total number of permutations containing two R’s.
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Last edited by JaneFairfax (2008-02-06 01:39:47)

goed

Member

Registered: 2008-02-05

Posts: 4

Re: Permutations and Combinations

Thank you for your quick reply.....but the answers are 1596(for selecting 4 letters from the given word) and 198 (containing 2 R's)and no clue how he gets that?? Waiting for your reply

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Permutations and Combinations

I made a slight mistake in my original working. I have edited my post and corrected it now. See my edited post. It should be correct now.

goed

Member

Registered: 2008-02-05

Posts: 4

Re: Permutations and Combinations

Thanks again,
I got the first one but the second one is a bit confusing.......but  4c2 = 4!/2!.2! = 6 then the final answer is 6x6p2 + 6x4c2 = 6x30 + 6x6 = 180+36 = 216.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Permutations and Combinations

Sorry, I meant to write 6×[sup]3[/sup]C[sub]2[/sub].
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