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solve for n;
a) C(n,2) = 28
b) P(n,5) = 20P(n,3)

2007-04-19 16:59:15 · 7 answers · asked by cutie pie! 3 in Science & Mathematics Mathematics

7 answers

Hi,

a) C(n,2) = 28 is true when n = 8

8 nCr 2 = 8!/(6! 2!) = 8*7/2 = 4 * 7 = 28


b) P(n,5) = 20P(n,3) is also true when n = 8

8nPr 5 = 20 * 8 nPr 3 ==>

8nPr 5 = 8!/3! = 8 x 7 x 6 x 5 x 4 = 6720

8nPr3 = 8 x 7 x 6 = 336, so 20 * 336 = 6720

Therefore, 8nPr 5 = 20 * 8 nPr 3


I hope that helps and shows why n = 8 for both of them.

2007-04-19 17:11:02 · answer #1 · answered by Pi R Squared 7 · 0 0

By definition
C(n,k) = n!/(k! (n - k)!)

Also, n! can be defined recursively, as
n! = n(n - 1)!

1) C(n,2) = 28

n! / (2! (n - 2)!) = 28

(n(n - 1)(n - 2)!) / ( 2! (n - 2)! ) = 28

n(n - 1) / 2! = 28
n(n - 1)/2 = 28
n(n - 1) = 56
n^2 - n = 56
n^2 - n - 56 = 0
(n - 8)(n + 1) = 0

Therefore, n = {8, -1}

However, n cannot be negative; therefore,

n = 8

b) P(n, r) = n! / (n - r)!

P(n, 5) = 20P(n, 3)

n! / (n - 5)! = 20 n!/(n - 3)!

Cross multiply,

n! (n - 3)! = 20 n! (n - 5)!

Divide both sides by n!,

(n - 3)! = 20 (n - 5)!

Express (n - 3)! recursively.

(n - 3)(n - 4)(n - 5)! = 20(n - 5)!

Divide both sides by (n - 5)!,

(n - 3)(n - 4) = 20

Solve for n.

n^2 - 7n + 12 = 20
n^2 - 7n - 8 = 0
(n - 8)(n + 1) = 0

Therefore,

n = {8, -1}

However, we reject n = -1 because n cannot be negative. Therefore,

n = 8.

2007-04-19 17:06:17 · answer #2 · answered by Puggy 7 · 0 0

a) C(n,2) = n! / [2!*(n-2)!] =n(n-1)/2=28
n(n-1)=56=8*7
n=8

b) P(n,5) = n! / (n-5)! = n(n-1)(n-2)(n-3)(n-4)
P(n,3) = n! / (n-3)! = n(n-1)(n-2)
P(n,5) = 20P(n,3)
n(n-1)(n-2)(n-3)(n-4) = 20*n(n-1)(n-2)
(n-3)(n-4)=20=5*4
n-3=5
n=8

2007-04-19 17:08:52 · answer #3 · answered by Anonymous · 0 0

a) n!/(n-2)!/2 = 28
=> n*(n-1) = 56
n^2 - n - 56 = 0

n = (1+ sqrt(1+4*56))/2 = (1+15)/2 = 8

b) n!/(n-5)! = 20 n!/(n-3)!
(n-4)(n-3)(n-2)(n-1)n = 20 (n-2)(n-1)n
=> (n-4)(n-3) = 20
=> n^2 -7n +12 - 20 = 0
n^2 - 7n -8 = 0

n = (7 + sqrt(49 +32))/2 = (7 + 9 )/2 = 8
n = 8

2007-04-19 17:07:54 · answer #4 · answered by roman_king1 4 · 0 0

a)

n(n - 1)/(2*1) = 28

n^2 - n = 56

n^2 - n - 56 = 0

(n - 8)(n + 7) = 0

n = 8, -7; discard -7


b)

n(n - 1)(n - 2)(n - 3)(n - 4) = 20*n(n - 1)(n - 2)

(n - 3)(n - 4) = 20

n^2 - 7n + 12 = 20

n^2 - 7n - 8 = 0

(n - 8)(n + 1) = 0

n = 8, -1; discard -1

2007-04-19 17:14:29 · answer #5 · answered by sweetwater 7 · 0 0

a) n=8
b)n=8

2007-04-19 17:07:04 · answer #6 · answered by bruinfan 7 · 0 0

Shut up and go to bed.

2007-04-19 17:03:26 · answer #7 · answered by Snowblind 2 · 0 0

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