If the letters of word PARIS are rearranged to form new words

Question

If the letters of word PARIS are rearranged to form new words, how many of these words have both P ahead of R and A ahead of S?

(A) 5
(B) 6
(C) 10
(D) 15
(E) 30

chetan2u · Accepted Answer

Apart from the solutions mentioned below...

Such questions are best done with combination formulas..

1) 5 letters so 5! ways
P and R can be arranged in 2! ways but ONLY one will have the given restriction
A and D also in 2! ways but ONLY one will have the given restriction.
Total -  \frac{5!}{2!2!}=5*3*2=30

2) P and R can choose two seats out of 5 in 5C3 so 10 ways
In the remaining 3, all three can be placed in 3! But half will have A ahead of D and half will have D ahead of A...so 3!/2=3 ways
Total 10*3=30 ways

E

souvonik2k · Answer

Total no. of words that can be formed from PARIS = 5! = 120.
P ahead of R will be in 60 words and R ahead of P will be in 60 words.
Out of these 60 words, 30 will contain A ahead of S and 30 will have S ahead of A.
Answer (E) 30.

vmmMBA · Answer

This is a Combinatorics problem! We are provided with a five letter word and is asked to find the total number of combinations where P is ahead of R and A is ahead of S.

First set of combinations : Here P and A takes the first two positions, and the remaining three letters takes any of the three positions

1 * 1 * 3 * 2 * 1  = 6 . But note that the positions of P and A can be interchanged, and still meet the condition.  So total words from the first set of combinations is 6*2=12

Second set of combinations : Here P (or A) takes the first position and A(or P) takes the third position. In the second position only R (or S if A has been placed in the first position) and I can be placed.

1 * 2 * 1 * 2 * 1  = 4 . But note that the positions of P and A can be interchanged, and still meet the condition.  So total words from the second set of combinations is 4*2=8

Third set of combinations : Here P (or A) takes the first position and A(or P) takes the fourth position. In the fifth position S (or R if P is in the fourth position) must be placed.
 
  1 * 2 * 1 * 1 * 1  = 2 . But note that the positions of P and A can be interchanged, and still meet the condition.  So total words from the third set of combinations is 2*2=4

Fourth set of combinations : Here P (or A) takes the second position and A(or P) takes the third position. Only I can take the first place.

1 * 1 * 1 * 2 * 1  = 2 . But note that the positions of P and A can be interchanged, and still meet the condition.  So total words from the fourth set of combinations is 2*2=4

Fifth set of combinations : Here P (or A) takes the second position and A(or P) takes the fourth position. Only I can take the first place. Only R (or S if A is in the second place) can take the third place.

1 * 1 * 1 * 1 * 1  = 1 . But note that the positions of P and A can be interchanged, and still meet the condition.  So total words from the fifth set of combinations is 1*2=2

Total combinations = 12 + 8 + 4 + 4 + 2 = 30

Answer is E!

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pushpitkc · Answer

We can use the answer options in order to arrive at the correct option in this case
If P has to be ahead of R and A ahead of S, some combinations that are possible:

AP--- The remaining 3 letters can be arranged in 3*2*1 or 6 ways
PA--- The remaining 3 letters can be arranged in 3*2*1 or 6 ways
PRA-- The remaining 2 letters can be arranged in 2*1 or 2 ways
PAR-- The remaining 2 letters can be arranged in 2*1 or 2 ways
PAS-- The remaining 2 letters can be arranged in 2*1 or 2 ways

The total number of ways is already greater than 15
Therefore, the only valid answer option that remains is  30(Option E)

GMATGuruNY · Answer

The total number of ways to arrange the 5 letters = 5! = 120.

The probability that P is ahead of R is equal to the probability that R is ahead of P.
Implication:
In  \frac{1}{2}  of the 120 arrangements above, P will be ahead of R:
 \frac{1}{2}*120 = 60

The probability that A is ahead of S is equal to the probability that S is ahead of A.
Implication:
In  \frac{1}{2}  of the 60 arrangements above, A will be ahead of S:
 \frac{1}{2}*60 = 30

E

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If the letters of word PARIS are rearranged to form new words

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