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) (n2)!  n(n2)! / n!
b) (nC2)  (nCn2)
4)If nPr  506 and (nCr_ = 253, find n and r.
5) If (28C2r)/(24C2r4) = 225/11, find r.

Data management  MathMate, Friday, October 28, 2011 at 12:14am
a)
P(2n,3)=2P(n,4)
=>
2n(2n1)(2n2)=2n(n1)(n2)(n3)
cancel 2n to get
(2n1)(2n2)=(n1)(n2)(n3)
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(n2)!n(n2)!]/n!
(n2)![n²n]/n!
=(n2)!n[n1]/n!
=n!/n!
=1
Give a try to 3b.
4.
What it is saying is that
P(n,r)=2C(n,r)
=>
n!/(nr)! = 2*n!/((nr)!r!)
Cancel the n! and (nr)! 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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