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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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Senior CR Moderator
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Updated on: 28 Nov 2016, 06:38
7
00:00

Difficulty:

65% (hard)

Question Stats:

50% (02:05) correct 50% (02:03) wrong based on 75 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

_________________

Originally posted by broall on 28 Nov 2016, 05:07.
Last edited by broall on 28 Nov 2016, 06:38, 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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28 Nov 2016, 09:28
7
3
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.

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

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

$$x + y + z + t = 30$$

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Karishma
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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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28 Nov 2016, 06:32
Top Contributor
1
(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
Hope it helps!

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

_________________

Ankit

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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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29 Nov 2017, 08:19
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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, 08:19
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