# QR probability question

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2

#### 299678

How many ways can we give 3 prizes to two student, when both of them equally eligible for all prizes?

solution manual said
A B
3 0
2 1
1 2
0 3
So total 4 possible ways

but, if I use combination method nCr, I only have 3 possible ways
why I can't use combination/or permutation method to solve this problem?

?

The numbers are so low, just do it manually ( I don't get their method...)

Prize 1,2,3 given to two people, A and B=

A-1
B-2

A-1
B-3

A-2
B-1

A-2
B-3

A-3
B-2

A-3
B-1

I don't understand why the answer isn't 6 because if person A gets a prize X and person B gets prize Y, that is not the same as person A getting prize Y and person B getting prize X? Shouldn't this be a permutation with no repetitions (http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html)

n!/ (n-r)!= (3!)/[(3-1)!]= 6

I think they mean that all 3 prizes are the same but the students can get a different amount of prizes.

So either

3 and 0
2 and 1
1 and 2
0 and 3

Terribly worded question. I'd usually go with the above answer of 6 from (3 P 2).

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I think they mean that all 3 prizes are the same but the students can get a different amount of prizes.

So either

3 and 0
2 and 1
1 and 2
0 and 3

Terribly worded question. I'd usually go with the above answer of 6 from (3 P 2).

Ah.....I gotcha. Didn't even make sense before you said they were all the same prize. Definitely terrible wording.

Ah.....I gotcha. Didn't even make sense before you said they were all the same prize. Definitely terrible wording.
The question is worded absolutely horribly. At first when I read it I thought someone translated it from a different language or summed up the question and primary language was not english (not joking)

Its a maximum of 3 awards, so you cant do 3, 1 etc

So if student a has...
0 awards
1 award
2 awards
3 awards

Student B much have
3 awards
2 awards
1 award
0 awards

There shouldnt be any combinations, therefore the answer would be 4.

What were you typing into your calculator when you were doing the ncr stuff.
And like someone else had said, its much easier with low numbers just to write it out.

How many ways can we give 3 prizes to two student, when both of them equally eligible for all prizes?

solution manual said
A B
3 0
2 1
1 2
0 3
So total 4 possible ways

but, if I use combination method nCr, I only have 3 possible ways
why I can't use combination/or permutation method to solve this problem?

For the First person, he can get prize#1 or prize#2 or prize#3, so there are 3 ways the first person can get a prize.

After the First person gets a prize, there are only 2 prizes remaining that could be awarded to the Second person.

How many ways can 2 people receive 3 prizes? 3 x 2 = 6 possible ways.

I guess you didn't read every other post.