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Given set M = {x natural/ (x+7)C(x+3) = 3(x+5)A2}, M is: A. M = nul set B M include in...

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yapayapa | Honors

Posted July 12, 2013 at 4:10 PM via web

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Given set M = {x natural/ (x+7)C(x+3) = 3(x+5)A2}, M is:

A. M = nul set

B M include in (2,5]

c. M include in (5,10]

d. M include [0,2]

Tagged with include, math, set

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 12, 2013 at 5:11 PM (Answer #1)

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You need to use the following factorial formulas, such that:

`C_n^k = (n!)/(k!(n-k)!)`

`A_n^k = (n!)/((n-k)!)`

Reasoning by analogy yields:

`C_(x+7)^(x+3) = ((x+7)!)/((x+3)!(x+7-x-3)!)`

`C_(x+7)^(x+3) = ((x+7)!)/((x+3)!(4)!) => C_(x+7)^(x+3) = ((x+3)!(x+4)(x+5)(x+6)(x+7))/((x+3)!(4)!)`

Reducing duplicate factors yields:

`C_(x+7)^(x+3) =((x+4)(x+5)(x+6)(x+7))/(4!) `

`A_(x+5)^2 = ((x+5)!)/((x + 5 - 2)!)`

`A_(x+5)^2 = ((x+5)!)/((x + 3)!)`

`A_(x+5)^2 = ((x+3)!(x+4)(x+5))/((x + 3)!)`

Reducing duplicate factors yields:

`A_(x+5)^2 = (x+4)(x+5)`

Replacing `((x+4)(x+5)(x+6)(x+7))/(4!)`   for `C_(x+7)^(x+3)`   and `(x+4)(x+5)` for `A_(x+5)^2` yields:

`((x+4)(x+5)(x+6)(x+7))/(4!) = 3(x+4)(x+5)`

Reducing duplicate factors yields:

`((x+6)(x+7))/24 = 3 => x^2 + 13x + 42 - 72 = 0`

`x^2 + 13x - 30 = 0`

Using quadratic formula yields:

`x_(1,2) = (-13+-sqrt(169 + 120))/2`

`x_(1,2) = (-13+-17)/2 => x_1 = 2 ; x_2 = -15`

Since the problem provides the information that the elements of the set `M` need to be natural numbers, you need to only keep the natural solution `x = 2` .

Hence, since the set ` M` contains the element `x = 2,` you need to select the answer d. `M` `sub` `[0,2]` .

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