6!/2!3!
For this problem, I did 6*5*4*3*2*1/2*1*3*1 for the answer 729/6=120
No , you messed up the denominator.
I will assume you meant 6!/(2!3!)
= 6*5*4*3*2*1/(2*1*3*2*1)
= 720/(2*6) = 720/12 = 60
And for c(8,6),
I did 8!/(8-6!) =8!/2! = 8*7*6*5*4*3*2*1/2*1 =40320/2=20160 but I think it is wrong.
C(n,r) = n!/[ r!(n-r)! ]
= 8! / [6!(2!)] = 8*7 / 2 = 4*7 = 28
To solve the expression 6!/2!3!, you need to evaluate the factorials and perform the division.
First, let's understand what factorials are. The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.
In this case, we have 6! in the numerator and 2!3! in the denominator.
To find the value of 6!, you multiply 6 by all the positive integers smaller than 6 until you reach 1:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
The value of 2! is 2, as it is the product of all positive integers smaller than 2:
2! = 2 * 1 = 2
The value of 3! is found similarly:
3! = 3 * 2 * 1 = 6
Now, substituting these values back into the expression:
6!/2!3! = 720 / (2 * 6)
To simplify the expression, we can cancel out the common factor of 2 in the numerator and denominator:
6!/2!3! = 720 / (2 * 6) = 720 / 12
Finally, performing the division:
6!/2!3! = 720 / 12 = 60
Therefore, the simplified value of 6!/2!3! is 60.