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C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C

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C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C  [#permalink]

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New post Updated on: 30 Oct 2016, 06:23
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C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C(5,3)=C(5,x) and x≠3, what is the value of x?


(A) 0
(B) 1
(C) 2
(D) 4
(E) 5

Originally posted by interceptor77 on 30 Oct 2016, 05:01.
Last edited by Bunuel on 30 Oct 2016, 06:23, edited 1 time in total.
Renamed the topic.
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Re: C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C  [#permalink]

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New post 30 Oct 2016, 05:17
1
interceptor77 wrote:
C(m,n) = (m!)/ ((m-n)!n!) for nonnegatives integers m and n, m>=n. If C(5,3)=C(5,x) and x≠3, what is the value of x?


(A) 0
(B) 1
(C) 2
(D) 4
(E) 5


In Combinatorics,

nCr = nC(n-r)

So x = 5-3 =2

C


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Re: C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C  [#permalink]

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New post 06 Dec 2017, 05:30
2
interceptor77 wrote:
C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C(5,3)=C(5,x) and x≠3, what is the value of x?


(A) 0
(B) 1
(C) 2
(D) 4
(E) 5


The given expression should remind you that it is mCn i.e. selecting n out of total m items.

We know that mCn = mC(m-n) and hence 5C3 = 5C(5-3) = 5C2

Even if this doesn't strike you during the exam, simply substitute.

(m!)/((m-n)!n!) = 5!/3!*2! = 5!/(5-x)!*x!

(5-x)!*x! = 3!*2!

We see that x = 2

Answer (C)
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Re: C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C &nbs [#permalink] 06 Dec 2017, 05:30
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