Posted by Jane on Thursday, October 27, 2011 at 9:55pm.
a) 2nP3 = 2(nP4)
b)6(n+1P2) = 3nP2
2) If (nC12 = (nC8), find (nC17) and (22Cn)
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
cancel 2n to get
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:
Give a try to 3b.
What it is saying is that
n!/(n-r)! = 2*n!/((n-r)!r!)
Cancel the n! and (n-r)! to get
=> 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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