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How many ways can you choose 3 cards so the total is 15?

  • Thread starter Natasha1
  • Start date
  • #1
467
7

Homework Statement



The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are printed on individual cards.

How many ways can you choose three cards so that the total of the three numbers is 15?

The Attempt at a Solution



The different ways I can see are:

0, 9, 6
0, 8, 7
1, 9, 5
1, 8, 6
1, 5, 9
2, 9, 4
2, 8, 5
2, 7, 6
2, 6, 7
2, 4, 9
3, 8, 4
3, 7, 5
4, 9, 2
4, 8, 3
4, 6, 5

Is 15 different ways the answer?
 

Answers and Replies

  • #2
467
7
Is 15 the right answer though?
 
  • #3
294
0
Does the order of picking the cards matter?

Also, is there replacement?


The question seems a little ambiguous.
 
  • #4
1,065
10
Maybe you put it this way.
0+(15)
1+(14)
2+(13)
3+(12)
4+(11)
5+(10)
..
..
9+(6)
where in bracket is the sum of 2 numbers.
 
  • #5
881
40
Another, generalized way to solve this problem would be to use Multinomial Theorem. The answer you need would be the coefficient of x15 in

[itex](1+x+x^2+......+x^9)(1+x+x^2+......+x^9)(1+x+x^2+......+x^9)[/itex]

which is equal to

[itex](1+x+x^2+......+x^9)^{3}[/itex]

This can be applied to any similar problem with any kind of conditions on the numbers, and its easier than listing all the numbers out, especially when numbers get bigger :wink:

If there is no replacement, you need to subtract the cases where the power of x is same.
 
  • #6
467
7
It does not mention about replacement in the question.

I think the answer is 15. Could someone please just say yes or no? As I have attempted it...
 
  • #7
881
40
It does not mention about replacement in the question.

I think the answer is 15. Could someone please just say yes or no? As I have attempted it...
The answer is wrong. :smile:


Edit : Also, as Villyer asked, does the order of picking the cards matter? Your answer is wrong in either case, though...
 
Last edited:
  • #8
467
7
It is not mentioned in the question at all. My gut feeling is that it does matter...
 
  • #9
881
40
It is not mentioned in the question at all. My gut feeling is that it does matter...
Okay, try solving for both cases.

Case 1 : It matters.
Case 2 : It doesn't.

Lets say your initial solution was for Case 1.

Why your solution is wrong, is because you have
1, 9, 5
1, 8, 6
1, 5, 9
Now there needs to be a few more ways that consider 5,1,9 and 5,9,1 and 9,1,5 and 9,5,1. This can be done to each of the numbers you listed down, so that would result in lots more cases than you already have.


If you had solved it for Case 2, then the 1,9,5 and 1,5,9 would mean the same. So you would have to eliminate such numbers from your answer.


PS : the number of Case 1 has a special relation with Case 2, so just doing Case 2 will suffice, after which you can multiply the 'extra ways.'
 
  • #10
467
7
Ok so I am only going to look at the case 2, because I don't think it matters actually. I did say it did sorry in an earlier post, I actually meant it didn't.

So surely these are the only 15 ways if not, could someone at least give me one more?...

0, 9, 6
0, 8, 7
1, 9, 5
1, 8, 6
1, 5, 9
2, 9, 4
2, 8, 5
2, 7, 6
2, 6, 7
2, 4, 9
3, 8, 4
3, 7, 5
4, 9, 2
4, 8, 3
4, 6, 5
 
  • #11
881
40
Did you read the latter part of my previous post?? :confused:

If you had solved it for Case 2, then the 1,9,5 and 1,5,9 would mean the same. So you would have to eliminate such numbers from your answer.
 
  • #12
467
7
so for case 1, the answer is 15 and case 2 it is 13 or is it case 1 = 13 and case 2 = 15
 
  • #13
1,065
10
Why 5,1,9 not counted?
 
  • #14
467
7
Is Case 2 correct like this with 10 ways? Or have I forgotten any?

0, 9, 6
0, 8, 7
1, 9, 5
1, 8, 6
2, 9, 4
2, 8, 5
2, 7, 6
3, 8, 4
3, 7, 5
4, 6, 5

God knows how many there are for Case 1, I won't put an answer in the box, never mind.
 
  • #15
473
13
That looks fine, although you probably could have made it easier (on yourself) to check by presenting each three-card set in ascending order.

As for the case where the order of selection matters, you just have to consider how many arrangements of each of the selections you have here can be chosen in.
 
  • #16
467
7
0, 6, 9
0, 7, 8
1, 5, 9
1, 6, 8
2, 4, 9
2, 5, 8
2, 6, 7
3, 4, 8
3, 5, 7
4, 5, 6

so that's 10 ways for Case 2.

For Case 1 is it just a matter of 10 by 6 as each line has 6 permutations possible? Hence, Case 1 = 60 different ways
 
  • #17
473
13
That's it. And also I think that no replacement is implied by the question, so you're done.
 

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