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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