Find the following permutation or combination.
1. 6P4
2. 5C2
P(n,r)=n!/(n-r)!
C(n,r)=n!/r!(n-r)!
So p(6,4)=? ,C(5,2)=?
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
To find the permutation and combination for the given numbers, we will use the following formulas:
Permutation formula = nPr = n! / (n - r)!
Combination formula = nCr = n! / (r! * (n - r)!)
1. 6P4:
To find the permutation of 6P4, where 6 is the total number of items and 4 is the number of items selected, we will use the permutation formula.
nPr = n! / (n - r)!
6P4 = 6! / (6 - 4)!
Calculating the factorial:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
(6 - 4)! = 2! = 2 * 1 = 2
Substituting these values into the permutation formula:
6P4 = 720 / 2
6P4 = 360
Therefore, the permutation of 6P4 is 360.
2. 5C2:
To find the combination of 5C2, where 5 is the total number of items and 2 is the number of items selected, we will use the combination formula.
nCr = n! / (r! * (n - r)!)
5C2 = 5! / (2! * (5 - 2)!)
Calculating the factorial:
5! = 5 * 4 * 3 * 2 * 1 = 120
2! = 2 * 1 = 2
(5 - 2)! = 3! = 3 * 2 * 1 = 6
Substituting these values into the combination formula:
5C2 = 120 / (2 * 6)
5C2 = 120 / 12
5C2 = 10
Therefore, the combination of 5C2 is 10.