+0

+12
285
11
+1071

Old MacDonald has 5 chickens, 4 donkeys, and 7 emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)

Mar 22, 2020

#1
+1071
+8

I thought the answer was 83, but it was wrong.

Mar 22, 2020
#3
+2866
+1

Example:

How many ways can you pair up 2 apples, 3 bananas, and 4 oranges? (With the same restrictions in your problem).

We first find how many pairs include an apple:
2 * 3 = 6

2 * 4 = 8

We then find how many pairs include a banana that is already not taken by the apple pairs

3 * 4 = 12

No more cases! (3 choose 2 is 3)

6 + 8 + 12 = 26

CalculatorUser  Mar 22, 2020
#6
+1071
+7

Using your method, I still get 83.

SpongeBobRules24  Mar 22, 2020
edited by SpongeBobRules24  Mar 22, 2020
edited by SpongeBobRules24  Mar 22, 2020
#7
+497
+1

I'm not trying to be rude, but did you take the time to read through CU's solution? It was an example problem, not the actual problem itself. What CU intends to do here is teach you the steps of the problem, not just how to read the answer and enter it in some website. Please take the time to actually try to learn what he's teaching you here. Using his method, you would get the answer of 83, which you already said yourself is wrong.

jfan17  Mar 22, 2020
edited by jfan17  Mar 22, 2020
edited by jfan17  Mar 22, 2020
#8
+1071
+8

I feel dumb now.

SpongeBobRules24  Mar 22, 2020
#9
+1071
+9

SpongeBobRules24  Mar 22, 2020
edited by SpongeBobRules24  Mar 22, 2020
#10
+2866
0

u sure it's not 83...?

CalculatorUser  Mar 23, 2020
#11
+1071
+8

Yes, I tried that answer and it said that it was incorrect.

SpongeBobRules24  Mar 26, 2020