shovan85 wrote:1 says r balls arranged in n slots
Number of ways = P(r,n) = 210
Formula of permutation P(r,n) = (r!)/(r-n)!
210 = 2*3*5*7 (factorise)
= 5*6*7
= 7!/4!
So when can we get P(r,n) = 7!/4!
=> r!/(r-n)! = 7!/(7-3)!
=> r = 7 and n = 3
Thus u got a concrete value of n and r. So satisfied.
2 says C(n,r) = 35 = 5*7
C(n,r) = n!/[r!*(n-r)!] = 5*7
We can have n = 7 and r = 4 here to satisfy the above equation.
But we can also have n = 7 and r = 3 to satisfy the same.
Thus no concrete value of r is found as r = 3 and r = 4 both satisfies. Thus Insufficient.
IMO
A
I marked D because i never thought that there can be different value for the same Combination.
I want to know its in permutation and then the value of N and R will always be Unique, Or even in Permutation like Combination we can have different values for N and R and still have the same answer.
I am asking this because if in exam time i have the same question and then if i able to break down the permutation to one particular value then i don't have to check it with different values because i can we sure that there is no other value that can have the same answer at the end.
