P(n,r)=n!/(n-r)!
C(n,r)=n!/r!(n-r)!
So p(6,4)=? ,C(5,2)=?
Find the following permutation or combination.
1. 6P4
2. 5C2
2 answers
Just as a shortcut, for nPk, start at n and count down for k factors. For example,
9P3 = 9*8*7
for combinations, do the same, but divide by k factors counting up
12C4 = (12*11*10*9) / (1*2*3*4)
Note that this makes it clear why you never have to go more than n/2 factors
12C8 = (12*11*10*9*8*7*6*5) / (1*2*3*4*5*6*7*8)
But note that the 5*6*7*8 cancels out, so
12C8 = 12C4
so if k > n/2, nCk = nCn-k
9P3 = 9*8*7
for combinations, do the same, but divide by k factors counting up
12C4 = (12*11*10*9) / (1*2*3*4)
Note that this makes it clear why you never have to go more than n/2 factors
12C8 = (12*11*10*9*8*7*6*5) / (1*2*3*4*5*6*7*8)
But note that the 5*6*7*8 cancels out, so
12C8 = 12C4
so if k > n/2, nCk = nCn-k
2 answers