+0  
 
+7
526
19
avatar+2314 

Yann and Camille go to a restaurant. If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order if they refuse to order the same dish? (It does matter who orders what---Yann ordering chicken and Camille ordering fish is different from Yann ordering fish and Camille ordering chicken.)

 

I got 90, but I'm not sure.

 Mar 22, 2020
edited by CalTheGreat  Mar 22, 2020
edited by CalTheGreat  Mar 22, 2020
 #1
avatar+497 
+1

Not too sure about this one, but I'll give it my best shot.

 

We know that There are 10 items on the menu, so say Yann goes first to order. She has 10 options to choose from, simple enough. Next, camille has 9 options to choose from, because they refuse to order the same dish, meaning she has 1 less option to order from. Then, we multiply this result by 2, because we can switch Yann and Camille around; if Camille goes first instead of Yann. We then get:

\(10*9*2 = 180\) ways of ordering dishes

 Mar 22, 2020
edited by jfan17  Mar 22, 2020
 #2
avatar+2314 
+2

Really? I got 90........  I was just checking. I did this:

 

 Yann can order 10 different dishes. After he chooses, Camille has 9 choices left. 10 times 9 is 90. 

 Mar 22, 2020
 #3
avatar+497 
+2

Did you look at my steps? I got 10*9*2 because it matters in what order they select their dishes. You can switch around Yann and Camille; that was the rationale behind my multiplying by 2

jfan17  Mar 22, 2020
 #4
avatar+2314 
+2

I did. But if we switch things around, isn't it the same thing? That's what I'm confused about.

CalTheGreat  Mar 22, 2020
 #8
avatar+1070 
+8

This same question was asked, but they could choose the same dish. CPhill got the answer of 100, and not 200.

 

https://web2.0calc.com/questions/help-plz_112

SpongeBobRules24  Mar 22, 2020
 #9
avatar+497 
+2

That wasn't the same question. In that question, the two were allowed to order the same dish whereas in this, they aren't 

jfan17  Mar 22, 2020
 #10
avatar+1070 
+8

So then why is it not 200.

SpongeBobRules24  Mar 22, 2020
 #12
avatar+497 
+1

not too sure myself! I understand the 10 * 9 part, but what I dont' understand is why you aren't multiplying by 2 because you need to factor in where the two are switched since the order does matter.

10 * 9 would be calculating how many Yann and Camille could order a dish alone, right? Not vice versa. 

To that end, let me ask a quick question:

 

How many ways are there for Yann and Camille to order dishes?

jfan17  Mar 22, 2020
 #5
avatar+1070 
+8

Yann can order from 10 dishes, but Camille can only order 9 because she cannot choose the same dish as Yann. So 10*9=90

 Mar 22, 2020
 #7
avatar+497 
+1

Read the last part of the question. "Yann ordering chicken and Camille ordering fish is different from Yann ordering fish and Camille ordering chicken." That's why you have to multiply by two, otherwise you're undercounting the cases where their orders are swapped around. Multiplying 10* 9 only counts for the cases where Yann is picking from the 10 and Camille is picking from the 9. Consider the case where Camille picks from 10 and Yann picks from 9. That's my logic here

jfan17  Mar 22, 2020
 #11
avatar+1070 
+9

If you write out all the combinations, you will see.

SpongeBobRules24  Mar 22, 2020
 #13
avatar+1070 
+9

Write out all the combinations if there were 3 items on the menu.

SpongeBobRules24  Mar 22, 2020
 #14
avatar+497 
+2

it appears that you are correct spongebob! I think I see what I did wrong here nevermind :P

jfan17  Mar 22, 2020
 #6
avatar+2314 
+2

Thanks for confirming! jfan17, do you agree with what he did?

 Mar 22, 2020
 #15
avatar+532 
+3

To get around the problem about the order of who orders the dish first, we can do \({10 \choose 2}\). However, as jfan17 said, we can switch the dishes so that Camille has what Yann ordered and Yann has what Camille ordered, so we multiply that by 2. This gives us a total amount of \(\frac{10\cdot9}{2}\cdot2\), which gives us an answer of 90 ways.

 Mar 22, 2020
edited by MathCuber  Mar 22, 2020
 #16
avatar+497 
+1

That makes a lot more sense in the context that I was thinking about the problem!

jfan17  Mar 22, 2020
 #17
avatar+2856 
+1

nice discussion 

 Mar 22, 2020
 #18
avatar+2314 
+2

Yes! Definitely!!!

CalTheGreat  Mar 22, 2020
 #19
avatar+111456 
+1

Agree with , CU......nice discussion....

 

 

 

cool cool cool

CPhill  Mar 22, 2020

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