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.

Questions about the same concept:

probability-85993.html?highlight=divide+groupscombination-55369.html#p690842probability-88685.html#p669025combination-groups-and-that-stuff-85707.html#p642634sub-committee-86346.html?highlight=divide+groupscombinations-problems-95344.html#p733805Hope it helps.

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