1)SOLVE:

Question

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.

Answer 1

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.

Answer 2

np4=88x n-1 p2

Answer 3

the question is very important question but you may answer each question step by step please