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# From the word TRAMPLE 4 letters are taken. What is the

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Manager
Joined: 05 Oct 2005
Posts: 81
From the word TRAMPLE 4 letters are taken. What is the [#permalink]

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25 Oct 2005, 04:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

From the word TRAMPLE 4 letters are taken. What is the probability that the word TRAM is arranged?

A. 1/2
B. 1/3
C. 1/7
D. 1/28
E. 1/35

I believe that all of these choices are incorrect. The correct answer should be 1/840.

If a word is arranged, then order matters, and we need to find the permutation instead of the combination. In this case, the probability of the word TRAM being arranged should be:

1/P(7,4)

P(7,4) = 7!/(7-4)! = 7!/3! = 7 x 6 x5 x 4 = 840

Thus the answer should be 1/840.

What do you guys think?
Senior Manager
Joined: 15 Apr 2005
Posts: 415
Location: India, Chennai
Re: Question Collection 1.2, #39 [#permalink]

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25 Oct 2005, 04:51
I am no expert in Probablity. But I think 1/840 is correct.

Here is my working.

The probablity of choosing T is 1/7
The probablity of choosing R is 1/6
The probablity of choosing A is 1/5
The probablity of choosing M is 1/4

Hence (1/7)*(1/6)*(1/5)*(1/4) = 1/840
Manager
Joined: 05 Oct 2005
Posts: 81
Re: Question Collection 1.2, #39 [#permalink]

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25 Oct 2005, 04:53
krisrini wrote:
I am no expert in Probablity. But I think 1/840 is correct.

Here is my working.

The probablity of choosing T is 1/7
The probablity of choosing R is 1/6
The probablity of choosing A is 1/5
The probablity of choosing M is 1/4

Hence (1/7)*(1/6)*(1/5)*(1/4) = 1/840

I agree with you. However, the answer in the question collection takes that number and multiplies by 24 (incorrectly, in my opinion) to reach an answer of 1/35.
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany
Re: Question Collection 1.2, #39 [#permalink]

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25 Oct 2005, 05:29
BumblebeeMan wrote:
krisrini wrote:
I am no expert in Probablity. But I think 1/840 is correct.

Here is my working.

The probablity of choosing T is 1/7
The probablity of choosing R is 1/6
The probablity of choosing A is 1/5
The probablity of choosing M is 1/4

Hence (1/7)*(1/6)*(1/5)*(1/4) = 1/840

I agree with you. However, the answer in the question collection takes that number and multiplies by 24 (incorrectly, in my opinion) to reach an answer of 1/35.

but there are two steps. firstly the letters are taken out and then the letters are arranged. that means 4 letters from 7 is 7c4 is 35. then 4 letters form 4, namely RAMT, is 4c4 is 1. now the 4 letters RAMT have to be arranged so that the word TRAM comes up. there is only 1 way to do so. so its 1/35...
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Manager
Joined: 05 Oct 2005
Posts: 81

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25 Oct 2005, 05:43
Quote:
but there are two steps. firstly the letters are taken out and then the letters are arranged. that means 4 letters from 7 is 7c4 is 35. then 4 letters form 4, namely RAMT, is 4c4 is 1. now the 4 letters RAMT have to be arranged so that the word TRAM comes up. there is only 1 way to do so. so its 1/35...

I disagree. The letters can be selected using 7c4 (which equals 35). However, you then have to multply 35 by the number of ways in which the letters can be arranged, which is 4!

35 x 4! = 840
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

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25 Oct 2005, 05:57
BumblebeeMan wrote:
Quote:
but there are two steps. firstly the letters are taken out and then the letters are arranged. that means 4 letters from 7 is 7c4 is 35. then 4 letters form 4, namely RAMT, is 4c4 is 1. now the 4 letters RAMT have to be arranged so that the word TRAM comes up. there is only 1 way to do so. so its 1/35...

I disagree. The letters can be selected using 7c4 (which equals 35). However, you then have to multply 35 by the number of ways in which the letters can be arranged, which is 4!

35 x 4! = 840

hm? you are right
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If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Intern
Joined: 25 Jun 2005
Posts: 23
Location: Bay Area, CA

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08 Nov 2005, 19:35
BumblebeeMan wrote:
I disagree. The letters can be selected using 7c4 (which equals 35). However, you then have to multply 35 by the number of ways in which the letters can be arranged, which is 4!

35 x 4! = 840

There are 35 ways to select 4 letters from 7 letters. However, there is exactly one permutation of 4 letters that we are picking (TRAM) among the 4! possibilities.

So solution is 1/35.
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Senior Manager
Joined: 03 Nov 2005
Posts: 385
Location: Chicago, IL

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25 Nov 2005, 22:06
Well, since the order the letters are chosen matters, Possible number of outcomes is 7P4=840. P=FO/EO=1/840. I thought about it this way.
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Senior Manager
Joined: 03 Nov 2005
Posts: 385
Location: Chicago, IL

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25 Nov 2005, 22:12
dil66 wrote:
BumblebeeMan wrote:
I disagree. The letters can be selected using 7c4 (which equals 35). However, you then have to multply 35 by the number of ways in which the letters can be arranged, which is 4!

35 x 4! = 840

There are 35 ways to select 4 letters from 7 letters. However, there is exactly one permutation of 4 letters that we are picking (TRAM) among the 4! possibilities.

So solution is 1/35.

you can't find probability (Favorable number of outcomes/ possible number of outcomes) unless you have both numerator and denominator in either combination or permutation. if order matters in the numerator, so it does in the denominator. I think
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SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka

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26 Nov 2005, 00:14
it should be 1/7P4 = 1/840

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hey ya......

26 Nov 2005, 00:14
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