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08 Jul 2025, 09:28
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This one is very tricky. Better to guess and move on unless you are a combinotarian (Don't google it up pls, I made it). It's not worth trying as it will take more than 2 mins to decode.
Anna has six pets and six treats to give out: two identical apple slices and four different fruits — a banana, a carrot, a mango, and a pear. Normally, without any restrictions, she could hand out the treats in many ways, accounting for the fact that the two apple slices are identical. However, there are two important constraints: her rabbit won’t eat the mango, and her parrot won’t eat the banana. To handle this, we first consider all the possible ways she could give out the treats if there were no refusals. Then, we subtract the situations where the rabbit is given the mango or the parrot is given the banana. But some of these bad assignments get counted twice — specifically, when the rabbit gets the mango and the parrot gets the banana at the same time — so we add those cases back once. After adjusting for all the invalid combinations, we find that the number of valid, distinct ways Anna can distribute the treats to her pets is 252.
Anna has six pets and six treats to give out: two identical apple slices and four different fruits — a banana, a carrot, a mango, and a pear. Normally, without any restrictions, she could hand out the treats in many ways, accounting for the fact that the two apple slices are identical. However, there are two important constraints: her rabbit won’t eat the mango, and her parrot won’t eat the banana. To handle this, we first consider all the possible ways she could give out the treats if there were no refusals. Then, we subtract the situations where the rabbit is given the mango or the parrot is given the banana. But some of these bad assignments get counted twice — specifically, when the rabbit gets the mango and the parrot gets the banana at the same time — so we add those cases back once. After adjusting for all the invalid combinations, we find that the number of valid, distinct ways Anna can distribute the treats to her pets is 252.
08 Jul 2025, 09:29
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NOETS:
6 pets.
Each gets 1 treat
2 apple slices + one of the following
Banana, Carrot, Mango, pear
just to visualize it:
Total number of combinations without restrictions....
[6!][/2!] = 720/2 =360 possible outcomes
Now how many potential mano to rabbit or banana to parrot.
[5!][/2!] = (120)/2 = 60 (both rabbit and parrot)
Last how many where we have both!
[4!][/2!] = 12
360 - (60 Parrot) - (60 Rabbit ) = 240 - (12 both) = 252
6 pets.
Each gets 1 treat
2 apple slices + one of the following
Banana, Carrot, Mango, pear
just to visualize it:
| Banana | Mango | Carrot | Pear | apple 1 | apple 2 | ||
| P1 | |||||||
| P2 | |||||||
| P3 | |||||||
| P4 | |||||||
| Rabbit | xxxxx | ||||||
| Parrot | xxxxxx | ||||||
Total number of combinations without restrictions....
[6!][/2!] = 720/2 =360 possible outcomes
Now how many potential mano to rabbit or banana to parrot.
[5!][/2!] = (120)/2 = 60 (both rabbit and parrot)
Last how many where we have both!
[4!][/2!] = 12
360 - (60 Parrot) - (60 Rabbit ) = 240 - (12 both) = 252
08 Jul 2025, 09:53
Kudos
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Let the pets be P1, P2, P3, P4, P5, P6
The treats are: A (apple slice), A (apple slice), B (banana), C (carrot), M (mango), P (pear).
There are 6 treats and 6 pets, so each pet gets exactly one treat.
The two apple slices are identical.
Given condition:
The (Pr)rabbit does NOT get the mango, AND the (Pp)parrot does NOT get the banana.
Now lets solve:
First, let's assign the non-apple treats (B, C, M, P) and then the apples(A,A)
Total ways to distribute 6 distinct items to 6 distinct pets is 6!=720.
However, since two apple slices are identical, we divide by 2! for the apples: 6!/2!=720/2=360.
Let N be the total number of ways to distribute the treats (360).
Let A be the set where the rabbit gets the mango.
Let B be the set where the parrot gets the banana.
We want to find N−∣A∪B∣=N−(∣A∣+∣B∣−∣A∩B∣).
lets calculate N (Total ways without restrictions on who gets what):
We have 6 pets and 6 distinct positions for the treats (since pets are distinct).
The treats are (A, A, B, C, M, P).
The number of ways to arrange these treats for the pets is 6!/2!=720/2=360.
lets calculate |A| (Rabbit gets Mango):
If the rabbit gets the mango, assign M to Pr
Now we have 5 treats left (A, A, B, C, P) and 5 pets left.
Number of ways = 5!/2!=120/2=60.
lets calculate |B| (Parrot gets Banana):
If the parrot gets the banana, assign B to Pp
Now we have 5 treats left (A, A, C, M, P) and 5 pets left.
Number of ways = 5!/2!=120/2=60.
now calculate |A intersect B| (Rabbit gets Mango AND Parrot gets Banana):
Assign M to Pr and B to Pp
Now we have 4 treats left (A, A, C, P) and 4 pets left.
Number of ways = 4!/2!=24/2=12.
Therefore, number of valid ways = N−(∣A∣+∣B∣−∣A∩B∣)
=360−(60+60−12)
=360−(120−12)
=360−108
=252
The treats are: A (apple slice), A (apple slice), B (banana), C (carrot), M (mango), P (pear).
There are 6 treats and 6 pets, so each pet gets exactly one treat.
The two apple slices are identical.
Given condition:
The (Pr)rabbit does NOT get the mango, AND the (Pp)parrot does NOT get the banana.
Now lets solve:
First, let's assign the non-apple treats (B, C, M, P) and then the apples(A,A)
Total ways to distribute 6 distinct items to 6 distinct pets is 6!=720.
However, since two apple slices are identical, we divide by 2! for the apples: 6!/2!=720/2=360.
Let N be the total number of ways to distribute the treats (360).
Let A be the set where the rabbit gets the mango.
Let B be the set where the parrot gets the banana.
We want to find N−∣A∪B∣=N−(∣A∣+∣B∣−∣A∩B∣).
lets calculate N (Total ways without restrictions on who gets what):
We have 6 pets and 6 distinct positions for the treats (since pets are distinct).
The treats are (A, A, B, C, M, P).
The number of ways to arrange these treats for the pets is 6!/2!=720/2=360.
lets calculate |A| (Rabbit gets Mango):
If the rabbit gets the mango, assign M to Pr
Now we have 5 treats left (A, A, B, C, P) and 5 pets left.
Number of ways = 5!/2!=120/2=60.
lets calculate |B| (Parrot gets Banana):
If the parrot gets the banana, assign B to Pp
Now we have 5 treats left (A, A, C, M, P) and 5 pets left.
Number of ways = 5!/2!=120/2=60.
now calculate |A intersect B| (Rabbit gets Mango AND Parrot gets Banana):
Assign M to Pr and B to Pp
Now we have 4 treats left (A, A, C, P) and 4 pets left.
Number of ways = 4!/2!=24/2=12.
Therefore, number of valid ways = N−(∣A∣+∣B∣−∣A∩B∣)
=360−(60+60−12)
=360−(120−12)
=360−108
=252
