If exactly 3 of the following 5 letters - A,B,B,C,C – are chosen to

Question

If exactly 3 of the following 5 letters - A,B,B,C,C – are chosen to create unique 3 letter arrangements in a row, how many such arrangements are possible?

A. 15
B. 18
C. 20
D. 30
E. 60

GMATinsight · Accepted Answer

Case 1 : 3 letters to be used are A, B, B 
Arrangements = 3!/2! = 3

Case 2 : 3 letters to be used are A, B, C 
Arrangements = 3! = 6

Case 3 : 3 letters to be used are A, C, C 
Arrangements = 3!/2! = 3

Case 4 : 3 letters to be used are B, C, C 
Arrangements = 3!/2! = 3

Case 5 : 3 letters to be used are B, B, C 
Arrangements = 3!/2! = 3

Total Arrangements = 3+6+3+3+3 = 18

Answer: Option B

chetan2u · Answer

There are repetition of letters, so let's see the ways the arrangements can be done..
1) All three different...A,B,C
These 3 can be arranged in 3! ways ...
2) Two of one kind...
These 2 can be B or C..
Each can be arranged with the remaining 2..
Finally each arrangement of 2 of one kind and 1 different... \frac{3!}{2}=3 
So ways =2*2*3=12 ways..

Total (1+2) =3!+12=6+12=18

B

guptarahul · Answer

There are repetition of letters
So we wil have cases ..

Case 1) All three different A,B,C
These 3 can be arranged in 3! ways = 6

Case 2) Two of one kind...
These 2 can be B or C, as A is only time present.
Each can be arranged with the remaining 2
Each arrangement of 2 of one kind and 1 different
Suppose first we choose A fix a's position
A _ _
A B B
C B B
A C C
B C C
2 2 1
So ways =2*2*3=12 ways..

Total (1+2) =3!+12=6+12=18

B

KarishmaB · Answer

Responding to a pm:

Actually, while picking the type of letters, some enumeration will be required.

The 3 letter chosen could be all distinct or 2 could be identical and 1 different. All 3 identical are not possible since there is no letter available thrice.

All 3 distinct: There is only 1 way of selecting them: A, B and C.
They can be arranged in a row in 3! ways.
Total = 1*3! = 6 ways

2 Same 1 different: The same letter can be chosen in 2 ways. The third letter can be chosen in 2 ways of the remaining two distinct letters. This gives us 4 ways of selecting the 3 letters in which 2 are identical and 1 is different.
They can be arranged in 3!/2! ways (because 2 are identical)
Total = 4*3!/2! = 12 ways

Hence total number of ways = 6 + 12 = 18

Answer (B)

JeffTargetTestPrep · Answer

We could list the possible arrangements. One such set, using only the letters ABB, would be: ABB, BAB, BBA. But listing all of the possibilities for all the letter combinations is time-consuming, so instead we can use the formula for indistinguishable permutations for sets of letter where a letter is repeated. (Otherwise, we will use the regular permutation formula.) Our possible arrangements are:

ABB = 3!/2! = 3 ways

ACC = 3!/2! = 3 ways

ABC = 3! = 6 ways

BBC = 3!/2! = 3 ways

BCC = 3!/2! = 3 ways

The total number of ways is 3 + 3 + 6 + 3 + 3 = 18 arrangements are possible.

Answer: B

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