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08 Jul 2025, 18:57
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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?
AABCMP: Available Treats;
Let's freeze six pets in a row so that the treat combination has only to be done.
The Total Way in which AABCMP can be arranged: 6!/2! = 360
Now, we have to remove the constraint cases:
Her rabbit refuses to eat the mango: All the cases where the rabbit was given mango(Freezing Mango for rabbit): 1 * 5!/2! = 60
Her parrot refuses to eat the banana: All the cases where the parrot was given Banana(Freezing banana for parrot): 1 * 5!/2! = 60;
Now, since these two constraints are independent, there are chances that these two occur simultaneously, and which may have been calculated twice, so we have to remove one instance of those: 1 * 1 * 4!/2 = 12
Final distribution: 360 - (60+60-12) = 252
(A) 96
(B) 240
(C) 252
(D) 288
(E) 300
AABCMP: Available Treats;
Let's freeze six pets in a row so that the treat combination has only to be done.
The Total Way in which AABCMP can be arranged: 6!/2! = 360
Now, we have to remove the constraint cases:
Her rabbit refuses to eat the mango: All the cases where the rabbit was given mango(Freezing Mango for rabbit): 1 * 5!/2! = 60
Her parrot refuses to eat the banana: All the cases where the parrot was given Banana(Freezing banana for parrot): 1 * 5!/2! = 60;
Now, since these two constraints are independent, there are chances that these two occur simultaneously, and which may have been calculated twice, so we have to remove one instance of those: 1 * 1 * 4!/2 = 12
Final distribution: 360 - (60+60-12) = 252
(A) 96
(B) 240
(C) 252
(D) 288
(E) 300
08 Jul 2025, 19:17
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Bunuel
It isn't easy to calculate multiple cases, Instead, we calculate the total and subtract the unfavorable cases.
Here we have to distribute 6 items of which 2 are same, So #ways = 6!/2! => 360.
If Rabbit got a mango :
We can distribute the others in any way, So 5!/2! => 60 ways.
Similarly, if a parrot got a banana: 60 ways.
But there might be a condition that we have added twice, when both parrot & rabbit got these 2. Hence 4!/2 => 12.
Total favorable ways = 360 - 60 -60 +12 => 252.
Hence IMO C
08 Jul 2025, 20:43
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Instead of finding all possible patterns with the given restrictions ,we can find the combinations with no restrictioons and subtract from the total combinations.
total combinations - 6 fruits distributed amongst 6 pets, with 2 identical apples
= \(\frac{6!} {2!} \)
= 360
Finding combinations with no restrictions
= no restriction with mango +no restriction with banana - no restriction with manga AND banana
no restriction with mango -> if rabbit eats mango, there are 5 fruits for the rest of the pets, with 2 identical apples
= \(\frac{5!}{2!}\)
= 60
no restriction with banana- if parrot eats banana, there are 5 fruits for the rest of the pets, with 2 identical apples
=\( \frac{5!}{2!}\)
= 60
no restriction with banana AND mango - if parrot eats banana and rabbit eats mango, there are 4 fruits for the rest of the pets, with 2 identical apples
= \(\frac{4!}{2!}\)
= 12
total combinations where restrictions would not be applied = 60+60-12 = 108
Now, combinations with restrictions = 360 - 108 = 252
Answer is C 252
total combinations - 6 fruits distributed amongst 6 pets, with 2 identical apples
= \(\frac{6!} {2!} \)
= 360
Finding combinations with no restrictions
= no restriction with mango +no restriction with banana - no restriction with manga AND banana
no restriction with mango -> if rabbit eats mango, there are 5 fruits for the rest of the pets, with 2 identical apples
= \(\frac{5!}{2!}\)
= 60
no restriction with banana- if parrot eats banana, there are 5 fruits for the rest of the pets, with 2 identical apples
=\( \frac{5!}{2!}\)
= 60
no restriction with banana AND mango - if parrot eats banana and rabbit eats mango, there are 4 fruits for the rest of the pets, with 2 identical apples
= \(\frac{4!}{2!}\)
= 12
total combinations where restrictions would not be applied = 60+60-12 = 108
Now, combinations with restrictions = 360 - 108 = 252
Answer is C 252
