Posted by **Jane** on Thursday, October 27, 2011 at 9:55pm.

1)SOLVE:

a) 2nP3 = 2(nP4)

b)6(n+1P2) = 3nP2

2) If (nC12 = (nC8), find (nC17) and (22Cn)

3)simplify.

a) n(squared) (n-2)! - n(n-2)! / n!

b) (nC2) - (nCn-2)

4)If nPr - 506 and (nCr_ = 253, find n and r.

5) If (28C2r)/(24C2r-4) = 225/11, find r.

- Data management -
**MathMate**, Friday, October 28, 2011 at 12:14am
a)

P(2n,3)=2P(n,4)

=>

2n(2n-1)(2n-2)=2n(n-1)(n-2)(n-3)

cancel 2n to get

(2n-1)(2n-2)=(n-1)(n-2)(n-3)

This kind of equation can be readily solved by trial and error, since they both increase monotonically at different rates.

In this case, n=1 (which is rejected) or n=8.

try (b) and (2)similarly to (a) above.

3(a) simplifies well, assuming you have left out the critical square brackets:

[n²(n-2)!-n(n-2)!]/n!

(n-2)![n²-n]/n!

=(n-2)!n[n-1]/n!

=n!/n!

=1

Give a try to 3b.

4.

What it is saying is that

P(n,r)=2C(n,r)

=>

n!/(n-r)! = 2*n!/((n-r)!r!)

Cancel the n! and (n-r)! to get

1=2/r!

=> r!=2 => r=2

After that, you only have to check by trial and error C(n,2)=253.

Give (5) a try.

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