If 40C(n,5)=C(n,3),then what is C(n,n-1) ?
1) 10
2) 20
3) 5
4) 6
To find the value of C(n, n-1), we first need to simplify the expression C(n,5) and C(n,3).
The formula for combinations, represented by C(n,r), is given by the formula: C(n,r) = n! / (r!(n-r)!), where n! represents the factorial of n.
Given that 40C(n,5) = C(n,3), let's break down both sides of the equation.
40C(n,5) = 40! / (5!(n-5)!)
C(n,3) = n! / (3!(n-3)!)
Now, let's simplify both expressions.
40! / (5!(n-5)!) = n! / (3!(n-3)!)
To further simplify the equation, we can cancel out some terms.
Since 40! contains factors of 5! and (n-5)!, and n! contains factors of 3! and (n-3)!, we can cancel them out.
(40! / 5!) / (n-5)! = (n! / 3!) / (n-3)!
After canceling out the factors, we have:
(40*39*38*37*36) / 5! = (n*(n-1)*(n-2)) / 3!
Now, let's simplify further.
(40*39*38*37*36) / (5*4*3*2*1) = (n*(n-1)*(n-2)) / (3*2*1)
Cancelling out the common factors, we have:
(40*39*38*37*36) = (n*(n-1)*(n-2))*5*4
Now, we can see that the left side of the equation is divisible by 5, 4, and 2 (since 40, 36, and the right side of the equation ends with 4). Therefore, the right side of the equation must also be divisible by 5, 4, and 2.
Thus, (n*(n-1)*(n-2)) must be divisible by 5, 4, and 2.
From the given options, only option 2) 20 satisfies this condition. Therefore, the value of C(n, n-1) is 20.