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# Different arrangements of set {A,B,A} if repetition is allowed

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Manager
Joined: 24 Nov 2016
Posts: 152
Different arrangements of set {A,B,A} if repetition is allowed  [#permalink]

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02 Aug 2018, 07:46
1
2
00:00

Difficulty:

65% (hard)

Question Stats:

34% (00:57) correct 66% (00:48) wrong based on 29 sessions

### HideShow timer Statistics

How many different arrangements of the set {A,B,A} are there, if repetition is allowed?
(ie. you could have BBB)

(A) 3
(B) 6
(C) 8
(D) 9
(E) 27

Source: NOVA GMAT Math Bible
Manager
Joined: 13 Feb 2018
Posts: 135
Re: Different arrangements of set {A,B,A} if repetition is allowed  [#permalink]

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02 Aug 2018, 07:54
With A,A,B 3!/2!=3
With B,B,A 3!/2!=3
With A,A,A 1
With B,B,B 1

Total 3+3+1+1=8

IMO
Ans: C
Manager
Joined: 24 Nov 2016
Posts: 152
Re: Different arrangements of set {A,B,A} if repetition is allowed  [#permalink]

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02 Aug 2018, 08:54
LevanKhukhunashvili wrote:
With A,A,B 3!/2!=3
With B,B,A 3!/2!=3
With A,A,A 1
With B,B,B 1

Total 3+3+1+1=8

IMO
Ans: C

Yep thats right. I'm wondering if there is a formula for the "permutations of indistinguishable objects with repetition allowed", but i looked everywhere with google and didn't find anything.

I came across one that said: $$\frac{n^r}{p!q!..z!}$$, but if you test it gives a different answer: $$\frac{3^3}{2!}=27/2≠8$$.
Re: Different arrangements of set {A,B,A} if repetition is allowed &nbs [#permalink] 02 Aug 2018, 08:54
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