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

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John has 5 friends who want to ride in his new car that can accommodat [#permalink]

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New post 04 Sep 2016, 09:30
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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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John has 5 friends who want to ride in his new car that can accommodat [#permalink]

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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]

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New post 17 Jan 2017, 06:37
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]

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New post 17 Jan 2017, 07:28
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]

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New post 22 Mar 2017, 13:20
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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New post 25 Mar 2017, 16:39
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]

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New post 25 Mar 2017, 20:21
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] 25 Mar 2017, 20:21
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