P(n,r)=(n!)/(r!(n-r)!)This would give you the number of possible permutations.n factorial over r factorial times n minus r factorial
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
The expression n2 - n - 56 factors to (n - 8)(n + 7).
n p =n!/(n-r)! r and n c =n!/r!(n-r)! r
The value of 6 minus n depends on the value of n and cannot be answered until known. Once n is known, subtract it from 6 to get your answer.
P(n,r)=(n!)/(r!(n-r)!)This would give you the number of possible permutations.n factorial over r factorial times n minus r factorial
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
It is 5n + 9r.
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
Algebraically it is written as ' n^(2) - n'. This factors to ' n(n-1)'.
The expression n2 - n - 56 factors to (n - 8)(n + 7).
2-n-7n = 2-8n
n minus five five subtracted from n
n p =n!/(n-r)! r and n c =n!/r!(n-r)! r
The value of 6 minus n depends on the value of n and cannot be answered until known. Once n is known, subtract it from 6 to get your answer.
n - 10 + 9n - 3 = 10n - 13
8n + 7 - n - 98 = 7n - 91