Homework Help

Calculate the binomial coefficient: 16C2 Use the Binomial Theorem to expand and...

user profile pic

idontlikeschool0 | Student, Grade 11 | (Level 1) Honors

Posted December 29, 2013 at 1:36 AM via web

dislike 1 like

Calculate the binomial coefficient: 16C2

Use the Binomial Theorem to expand and simplify the expression (2x + 3y)4

Expand the binomial by using Pascal's triangle to determine the coefficents (2x - 4y)6

Find the specified nth term in the expansion of the binomial. (Write the expansion in descending powers of x.)
(x + 5y)10, n = 4

Find the specified nth term in the expansion of the binomial. (Write the expansion in descending powers of x.)
(x - 4y)6, n = 3

Evaluate using a graphing utility: 15P5

Solve for n. n-1P2=n+1P3

You are given the probability that an event will happen. Find the probability that the event will not happen.
   P(E) = 0.45

2 Answers | Add Yours

user profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted December 29, 2013 at 6:24 AM (Answer #1)

dislike 1 like

The binomial coefficient 16C2 has to be calculated.

`nCr=(n!)/(r!*(n-r)!)`

Hence, `16C2=(16!)/(2!*(16-2)!)`

`=(16*15)/2`

`=120`

Using binomial theorem the expansion of `(2x+3y)^4` is:

`=4C0*(2x)^4*(3y)^0+4C1*(2x)^3*(3y)^1+4C2*(2x)^2*(3y)^2+4C3*(2x)^1*(3y)^3+4C4*(2x)^0*(3y)^4`

`=16x^4+4*8x^3*3y+6*4x^2*9y^2+4*2x*27y^3+81y^4`

`=16x^4+96x^3y+216x^2y^2+216xy^3+81y^4`

The binomial `(2x-4y)^6 ` has to be expanded using Pascal’s Triangle. The binomial coefficients from the sixth row of Pascal’s Triangle are: 1,6,15,20,15,6,1.

So, the expansion of `(2x-4y)^6` is:

`=1*(2x)^6-6*(2x)^5*(4y)+15*(2x)^4*(4y)^2-20*(2x)^3*(4y)^3+15*(2x)^2*(4y)^4-6*(2x)*(4y)^5+1*(4y)^6`

`=64x^6-768x^5y+3840x^4y^2-10240x^3y^3+15360x^2y^4-12288xy^5+4096y^6`

From the binomial theorem `(r+1)th` term is `nCrx^(n-r)y^r` . So, r is one less than the number of the term you need. So, to find the fourth term in the binomial expansion` (x+5y)^10` , use `r=3` .

So, `10C3*x^(10-3)*(5y)^3=120*x^7*125y^3=15000x^7y^3.`

For the expansion, refer to the method as shown above.

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 29, 2013 at 3:31 PM (Answer #2)

dislike 1 like

You need to find the n-th term of the binomial expansion `(x - 4y)^6,` for `n = 3` , such that:

`(x - 4y)^6 = C_6^0*x^6*(-4y)^0 + C_6^1*x^5*(-4y)^1 + C_6^2*x^4*(-4y)^2 + ...`

Replacing the binomial coefficients `C_6^0, C_6^1, C_6^2` by their values, yields:

`C_6^0 = 1`

`C_6^1 = 6`

`C_6^2 = (6(6-1))/2 => C_6^2 = 15`

`(x - 4y)^6 = x^6 - 6x^5*4y + 15x^4*16y^2 + ...`

`(x - 4y)^6 = x^6 - 24x^5*y + 240x^4*y^2 + ...`

Hence, evaluating the third term of the given binomial expansion yields `T_3 = 240x^4*y^2` .

You need to solve for n the factorial equation `(n-1)P2 = (n+1)P3` , using factorial formula, such that:

`nPk = (n!)/((n - k)!)`

Reasoning by analogy, yields:

`(n-1)P2 = ((n-1)!)/((n-1-2)!) => (n-1)P2 = ((n-1)!)/((n-3)!)`

`(n+1)P3 = ((n+1)!)/((n+1-3)!) =>(n+1)P3 = ((n+1)!)/((n-2)!)`

`(n-1)P2 = (n+1)P3 => ((n-1)!)/((n-3)!) = ((n+1)!)/((n-2)!)`

`((n-1)!)/((n-3)!) = ((n-1)!*n*(n+1))/((n-3)!(n-2))`

Reducing duplicate factors yields:

`1 = n*(n+1)/(n-2) => n - 2 = n^2 + n => n^2 = -2` invalid

Hence, evaluating the solution to the given equation, yields that there exists no solutions.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes