Given: nc10 = nc14

To find: 25cn

Solution:

We know that:

ncr = n! / r! (n-r)!

Using this formula, we can simplify nc10 and nc14 as:

nc10 = n! / 10! (n-10)!

nc14 = n! / 14! (n-14)!

Multiplying both equations, we get:

nc10 * nc14 = n! / 10! (n-10)! * n! / 14! (n-14)!

Simplifying further, we get:

nc10 * nc14 = n!^2 / (10! * 14!) (n-10)! (n-14)!

Now, we need to find 25cn:

25cn = 25 * n! / (25 - n)! n!

Simplifying this equation, we get:

25cn = 25! / (25 - n)!

Since nc10 = nc14, we can equate the above two equations:

n!^2 / (10! * 14!) (n-10)! (n-14)! = 25! / (25 - n)!

Cross-multiplying, we get:

n!^2 * (25 - n)! = 10! * 14! * (n-10)! * (n-14)! * 25!

Now, we can simplify this equation using factorials and algebraic manipulation to get:

n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-15)(n-16)(n-17)(n-18)(n-19)(n-20)(n-21)(n-22)(n-23)(n-24) = 25 * 24^2 * 23 * 22 * 21 * 20 * 19 * 18 * 17 * 16 * 15

Dividing both sides by 24^2, we get:

n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-15)(n-16)(n-17)(n-18)(n-19)(n-20)(n-21)(n-22)(n-23)(n-24) = 25 * 23 * 22 * 21 * 20 * 19 * 18 * 17 * 16 * 15

Now, we can substitute n = 25 in the above equation to get:

25 * 24 * 23 * 22 * 21 * 20 * 19 * 18 * 17 * 16 * 15 = 25 * 23 * 22 * 21 * 20 * 19 * 18 * 17 * 16 * 15

Canceling out the common factors, we get:

25 * 24 = 24^2

Simplifying this equation, we get:

25 = 24

This is a contradiction, and hence, our assumption that nc10 = nc14 is incorrect.

Therefore, we cannot determine the value of 25cn.

Hence, the correct answer is option (B) 25.

24c10=24c14. Hence n=24. Therefore 25c24=25c1=25.
b is the answer.