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

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.

goyalsau wrote:

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.

This is not the case in Permutation. The prevalence of multiple options in Combinatorics is the presence of both r! and (n-r)! in the denominator. As these two values will always be complement to each other they will always have two values (if not mentioned otherwise)

Always true C(n,r) = C(n,n-r) .

Hope this helps