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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, 07:32

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 So, answer will be D. Hope it helps!

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, 09:19

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