jusjmkol740 wrote:

10 Arabian horses are split into pairs to pull one of the distinct 4 carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible?

a) 420

b) 1260

c) 5220

d) 9450

e) 113400

Note: Found this question. But not the OA. Can anyone help? Thanks in advance.

Note: The question is beyond the GMAT scope.# of ways 10 horses can be divided into 5 groups when order of the groups does not matter is: \(\frac{C^2_{10}*C^2_{8}*C^2_{6}*C^2_{4}*C^2_{2}}{5!}=945\).

So we would have 945 different cases of dividing 10 horses into 5 groups:

1. \(A_1\), \(B_1\), \(C_1\), \(D_1\), \(E_1\) (each letter represent pair of horses);

2. \(A_2\), \(B_2\), \(C_2\), \(D_2\), \(E_2\);

...

945. \(A_{945}\), \(B_{945}\), \(C_{945}\), \(D_{945}\), \(E_{945}\).

Now, we want to assign a pair (a letter in our case) to one of 4

distinct carts. For each case (out of 945) \(P^4_5=120\) would represent # of ways to choose 4 pairs out of 5 when order matters (as carts are distinct). OR \(C^4_5\) - choose 4 different letters out of 5 and \(4!\) - # of ways to assign 4 different letters to 4 distinct carts: \(C^4_5*4!=120\).

So total # of different assignments would be \(945*120=113400\).

Answer: E.

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