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