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

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If the letters of word PARIS are rearranged to form new words, how ma  [#permalink]

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New post 21 Jul 2018, 05:54
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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

New question !!!..

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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If the letters of word PARIS are rearranged to form new words, how ma  [#permalink]

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New post 21 Jul 2018, 05:58
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chetan2u wrote:
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

New question !!!..


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
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If the letters of word PARIS are rearranged to form new words, how ma  [#permalink]

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New post 21 Jul 2018, 07:00
1
chetan2u wrote:
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

New question !!!..


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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If the letters of word PARIS are rearranged to form new words, how ma  [#permalink]

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New post 21 Jul 2018, 07:05
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chetan2u wrote:
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

New question !!!..


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.
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Re: If the letters of word PARIS are rearranged to form new words, how ma  [#permalink]

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New post 21 Jul 2018, 12:23
chetan2u wrote:
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

New question !!!..


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)
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Re: If the letters of word PARIS are rearranged to form new words, how ma  [#permalink]

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New post 21 Jul 2018, 13:24
1
chetan2u wrote:
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


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:
\(\frac{1}{2}*60 = 30\)


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Re: If the letters of word PARIS are rearranged to form new words, how ma &nbs [#permalink] 21 Jul 2018, 13:24
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