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If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5

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If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5 [#permalink]

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New post Updated on: 28 Nov 2016, 07:38
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A
B
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D
E

Difficulty:

  65% (hard)

Question Stats:

54% (01:07) correct 46% (01:32) wrong based on 72 sessions

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If \((a+b)^5=\)

\(a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\)

What is the value of \(x+y+z+t\)?

A. 7
B. 15
C. 20
D. 30
E. 35

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Originally posted by broall on 28 Nov 2016, 06:07.
Last edited by broall on 28 Nov 2016, 07:38, edited 1 time in total.
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Re: If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5 [#permalink]

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New post 28 Nov 2016, 10:28
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nguyendinhtuong wrote:
If \((a+b)^5=\)

\(a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\)

What is the value of \(x+y+z+t\)?

A. 7
B. 15
C. 20
D. 30
E. 35


I will just put a = b = 1 here.

\((a + b)^5 = (1 + 1)^5 = 2^5 = 32\)

\(a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5 = 1 + x + y + z + t + 1 = 32\)

\(x + y + z + t = 30\)

Answer (D)
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Re: If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5 [#permalink]

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New post 28 Nov 2016, 07:32
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(a+b)^5 can be opened using binomial theorem
(a+b)^5 = (5C0)a^5b^0 + (5C1)a^4b^1 + (5C2)a^3b^2 + (5C3)a^2b^3 + (5C4)a^1b^4 + (5C5)a^0b^5

so, x + y + z + t = (5C1) + (5C2) + (5C3) + (5C4)
= 5!/(1!*4!) + 5!/(2!*3!) + 5!/(3!*2!) + 5!/(4!*1!)
= 5 + (5*4)/2 + (5*4)/2 + 5 = 30
So, answer will be D.
Hope it helps!

More About Binomial Theorem

nguyendinhtuong wrote:
If \((a+b)^5=\)

\(a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\)

What is the value of \(x+y+z+t\)?

A. 7
B. 15
C. 20
D. 30
E. 35

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Re: If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5 [#permalink]

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New post 29 Nov 2017, 09:19
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Re: If (a+b)^5 = a^5+x*a^4*b+y*a^3*b^2+z*a^2*b^3+t*a*b^4+b^5   [#permalink] 29 Nov 2017, 09:19
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