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08 Jul 2025, 08:00
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C
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Difficulty:
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56% (02:10) correct
44%
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wrong
based on 244
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Anna has six pets and plans to give each one a treat. The treats consist of two identical apple slices, along with one banana, one carrot, one mango, and one pear. Her rabbit refuses to eat the mango, and her parrot refuses to eat the banana. In how many distinct ways can she distribute the treats among the pets?
(A) 96
(B) 240
(C) 252
(D) 288
(E) 300
(A) 96
(B) 240
(C) 252
(D) 288
(E) 300
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08 Jul 2025, 08:30
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Bunuel
GMAT Club Official Explanation:
Step 1: Total arrangements without restrictions
The number of ways to distribute the treats A, A, B, C, M, and P among six pets without any restriction is 6!/2! = 360.
Step 2: Invalid cases where the rabbit gets mango or the parrot gets banana
We now count the number of invalid cases where the rabbit gets mango or the parrot gets banana.
Fix mango to the rabbit. The remaining 5 items, A, A, B, C, and P, can be assigned to the other 5 pets (including the parrot) in 5!/2! = 60 ways.
Fix banana to the parrot. The remaining 5 items, A, A, C, M, and P, can be assigned to the other 5 pets (including the rabbit) in 5!/2! = 60 ways.
However, both cases above include the overlap where the rabbit gets mango and the parrot gets banana. To avoid double-counting, we subtract those. Fix both: mango to the rabbit and banana to the parrot. The remaining 4 items, A, A, C, and P, can be assigned to the other 4 pets in 4!/2! = 12 ways.
This makes the number of invalid cases equal to 60 + 60 - 12 = 108.
Step 3: Subtract invalid cases from total
Therefore, valid cases = total arrangements - invalid cases = 360 - 108 = 252.
Answer: C.
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08 Jul 2025, 08:18
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* 6 pets total (including a rabbit and a parrot)
* 6 treats: 2 identical apple slices, 1 banana, 1 carrot, 1 mango, 1 pear
* Restrictions: rabbit can't have mango, parrot can't have banana
I'll approach this by calculating the total number of ways to distribute the treats, then subtract the invalid arrangements.
Without any restrictions, the number of ways to distribute the treats would be:
6!/(21) = 720/2 = 360
The division by 2l accounts for the fact that the two apple slices are identical.
Now I need to subtract the invalid arrangements:
Case 1: Rabbit gets mango (invalid)
If the rabbit gets the mango, we need to distribute the remaining 5 treats among the 5 remaining pets. This can be done in 5l/(21) = 60 ways.
Case 2: Parrot gets banana (invalid)
If the parrot gets the banana, we need to distribute the remaining 5 treats among the 5 remaining pets. This can be done in 5l(21) = 60 ways.
Case 3: Both invalid cases occur simultaneously (rabbit gets mango AND parrot gets banana)
We need to subtract this case once since we've double counted it. This can happen in 41(21) = 12 ways.
Therefore, the total number of valid arrangements is:
360 - 60 - 60 + 12 = 252
The answer is (C) 252.
* 6 treats: 2 identical apple slices, 1 banana, 1 carrot, 1 mango, 1 pear
* Restrictions: rabbit can't have mango, parrot can't have banana
I'll approach this by calculating the total number of ways to distribute the treats, then subtract the invalid arrangements.
Without any restrictions, the number of ways to distribute the treats would be:
6!/(21) = 720/2 = 360
The division by 2l accounts for the fact that the two apple slices are identical.
Now I need to subtract the invalid arrangements:
Case 1: Rabbit gets mango (invalid)
If the rabbit gets the mango, we need to distribute the remaining 5 treats among the 5 remaining pets. This can be done in 5l/(21) = 60 ways.
Case 2: Parrot gets banana (invalid)
If the parrot gets the banana, we need to distribute the remaining 5 treats among the 5 remaining pets. This can be done in 5l(21) = 60 ways.
Case 3: Both invalid cases occur simultaneously (rabbit gets mango AND parrot gets banana)
We need to subtract this case once since we've double counted it. This can happen in 41(21) = 12 ways.
Therefore, the total number of valid arrangements is:
360 - 60 - 60 + 12 = 252
The answer is (C) 252.
