Total 10 horses, need to pick 2 from 10 for a cart;

You have;

10C2 * 8C2 * 6C2 * 4C2 = 113400

This one is not clear. Some expert please help.

Thanks
Mankey

Ten Horses are split into pairs to pull one of the distinct four carts in a race. If each cart is assigned to a pair of horses, how many different assignments of horses to carts are possible?

Shankar's approach is the way I approached it as well.

When we choose a pair of horses to pull each cart, since order does not matter, it is a combination (rather than a permutation).

For the first cart, we have 10 horses to choose from, and we must choose 2. This is 10C2 = 10!/(8!2!) = 10*9/2 = 45.

For the second cart, we have only 8 horses left to choose from, and we must choose 2. This is 8C2 = 8!/(6!2!) = 8*7/2 = 28.

For the third cart, we have only 6 horses left to choose from, and we must choose 2. This is 6C2 = 6!/(4!2!) = 6*5/2 = 15.

And, for the fourth and final cart, we have only 4 horses left to choose from, and we must choose 2. This is 4C2 = 4!/(2!2!) = 4*3/2 = 12.

We do not need to do anything further for the last two horses. This covers all possible pairs of horses for each of the carts.

45*28*15*6 is an annoying thing to calculate.
I recommend pairing the math as follows:
45*6 = 270.
28*15 = 420.
270*420 = 113400.