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John has 5 friends who want to ride in his new car that can accommodat

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sarthaksabharwal
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John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  04 Sep 2016, 10:30
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John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A) 3
B) 8
C) 10
D) 15
E) 20
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abhimahna
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John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  04 Sep 2016, 10:36
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sarthaksabharwal wrote:
John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A) 3
B) 8
C) 10
D) 15
E) 20

5 Friends, We have to select two friends out of these 5(as the Car can accommodate only 3 passengers at a time, John + other 2),

So, 5C2 = 10

Hence, C
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John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  17 Jan 2017, 07:37
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I am not sure if - 5C2 is going to be correct one .... It seems to me that 5C3 is going to be more precise answer. Because we need to select group of 3 people out of 5. Though both of them will result in 10 . Please clarify of choosing 5C2 over 5C3.
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  17 Jan 2017, 08:28
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sarthaksabharwal wrote:
John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A) 3
B) 8
C) 10
D) 15
E) 20


\(5C_3 = 10\)

Answer will be (C) 10
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John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  22 Mar 2017, 14:20
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both 5c3 and 5c2 are same= 10

passenger includes the driver or not that is the big question here? I believe that is true based on the advertisements of SUV - example 07 seaters, means the driver is included here

so the answer should be 5c2
in case in future these numbers get change
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  25 Mar 2017, 17:39
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Use equation nCr = \(\frac{(n!)}{[(n-r)!r!]}\) to choose n objects taken r at a time.

5C3 = \(\frac{(5!)}{(2!3!)}\) = 5 x 2 = 10

Answer C. 10
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  25 Mar 2017, 21:21
tanmay056 wrote:
I am not sure if - 5C2 is going to be correct one .... It seems to me that 5C3 is going to be more precise answer. Because we need to select group of 3 people out of 5. Though both of them will result in 10 . Please clarify of choosing 5C2 over 5C3.


\(5C2 = 5C3\)

Note that \(C^k_n =\frac{n!}{k!(n-k)!}\)

\(C^{n-k}_n=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{(n-k)!k!}\)

Hence \(C^k_n = C^{n-k}_n\)

That's why \(C^3_5 = C^2_5\)

Now let's compare \(5C2\) vs \(5C3\).

If you choose 3 out of 5, there are \(5C3\) different ways.

If you choose 3 out of 5, that means you drop 2 of 5 and 3 left. Thus there are \(5C2\) different ways.

Two different ways with different concept, but have the same result. So you could choose any one which suits you.
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  26 Sep 2018, 12:07
broall wrote:
tanmay056 wrote:
I am not sure if - 5C2 is going to be correct one .... It seems to me that 5C3 is going to be more precise answer. Because we need to select group of 3 people out of 5. Though both of them will result in 10 . Please clarify of choosing 5C2 over 5C3.


\(5C2 = 5C3\)

Note that \(C^k_n =\frac{n!}{k!(n-k)!}\)

\(C^{n-k}_n=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{(n-k)!k!}\)

Hence \(C^k_n = C^{n-k}_n\)

That's why \(C^3_5 = C^2_5\)

Now let's compare \(5C2\) vs \(5C3\).

If you choose 3 out of 5, there are \(5C3\) different ways.

If you choose 3 out of 5, that means you drop 2 of 5 and 3 left. Thus there are \(5C2\) different ways.

Two different ways with different concept, but have the same result. So you could choose any one which suits you.


I've seen in other questions in which I would have to multiply Combinations or subtract an integer from the Combination. Can you tell me why here we only do 1 Combination?
I think when you need to multiply is when there are, for example, women and men. And to subtract I'm not sure of an instance at the moment.
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atr0038
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  28 Dec 2018, 23:11
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Does it really take people only 22 seconds on average to get this question correct? I'm lagging behind, because I took almost double that amount of time to read the question, process it, and do then math. It took me like 38 seconds to solve this, but how can I get as fast as the average person at solving these types of problems? There must be some kind of shortcut if that many people can solve it that quickly.
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nust2017
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  29 Dec 2018, 03:44
Can anyone please explain the solution to this question for someone with average quant skills. I took more than 1 minute to answer the question, and the average time taken for the question is 22 seconds :O
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  09 Jan 2022, 22:30
nust2017 wrote:
Can anyone please explain the solution to this question for someone with average quant skills. I took more than 1 minute to answer the question, and the average time taken for the question is 22 seconds :O


Dear nust2017

try to think in this way,

there are only 3 places: + + +
But 5 friends that signifies 2 friends without places: X X

Hence we can represent: + + + X X
That is 3! and 2! our denominator

Total 5 friends or 5! is a numerator

\(\frac{5!}{3!*2!}\) = 10

Hope it helps :)
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rxb266
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink] New post  28 Feb 2023, 02:58
The different combinations of 3 passengers that can be formed from the 5 friends:

5C3 = 5! / (3!x2!) = 10

option C
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Re: John has 5 friends who want to ride in his new car that can accommodat [#permalink]
28 Feb 2023, 02:58
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