I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Thursday, September 27, 2007
How Many Ways Can You Get to 100?
How many strictly positive integer solutions (x, y, z) are there, such that x + y + z = 100
my unofficial answer will be 5,151 combinations. Here's my thinking:
where x=0, y+z=100 | 101 combinations
where x=1, y+z=99 | 100 combinations
where x=2, y+z=98 | 99 combinations
therefore, x=n, y+z=101-n | 101-n combinations.
101 sum(x) = (101+1)(101/2)= 5151 x=0
I think that (but dont have the time to check) covers all the combinations because for each value of x, every possible value of y+z is calculated, therefore for every value of y and z, x+y and x+z = 100 respectively.
The answer is 4851 ways. It helps if you think of putting 100 stones into three piles.
For a technical answer, read on:
By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles.
Now, view the k piles as a numbered set .
Write on each stone the number of a chosen pile and order the stones accordingly.
The numbered stones constitute a combination with repetition of k elements (the numbers) choose m-k (the stones). The number of combinations can be calculated as: (m-1)! --------- (m-k)!(k-1)! In our case, m = 100 and k = 3, so we have 99!/97!*2! = 4851
well technically o is not positive or negative so u can not figure one of the numbers as zero and do decimals count? because if so then the answer would b infinite
Hi anonymous, I'm not sure I understand your question, so forgive me. But if each of the x, y and z need to be strictly positive integers, then 0 is out and so are decimals.
"By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles."
Hey there, this is a different annonymous user but when you say "possibly empty" is that erroneous considering you are leaving 0 out of the question. (Strictly positive integers)
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my unofficial answer will be 5,151 combinations. Here's my thinking:
ReplyDeletewhere x=0, y+z=100 | 101 combinations
where x=1, y+z=99 | 100 combinations
where x=2, y+z=98 | 99 combinations
therefore, x=n, y+z=101-n | 101-n combinations.
101
sum(x) = (101+1)(101/2)= 5151
x=0
I think that (but dont have the time to check) covers all the combinations because for each value of x, every possible value of y+z is calculated, therefore for every value of y and z, x+y and x+z = 100 respectively.
please correct me if my logic is wrong.
The answer is 4851 ways. It helps if you think of putting 100 stones into three piles.
ReplyDeleteFor a technical answer, read on:
By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles.
Now, view the k piles as a numbered set .
Write on each stone the number of a chosen pile and order the stones accordingly.
The numbered stones constitute a combination with repetition of k elements (the numbers) choose m-k (the stones). The number of combinations can be calculated as:
(m-1)!
---------
(m-k)!(k-1)!
In our case, m = 100 and k = 3, so we have 99!/97!*2! = 4851
Although, I could be wrong. ;-)
well technically o is not positive or negative so u can not figure one of the numbers as zero and do decimals count? because if so then the answer would b infinite
ReplyDeleteHi anonymous, I'm not sure I understand your question, so forgive me. But if each of the x, y and z need to be strictly positive integers, then 0 is out and so are decimals.
ReplyDelete"By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles."
ReplyDeleteHey there, this is a different annonymous user but when you say "possibly empty" is that erroneous considering you are leaving 0 out of the question. (Strictly positive integers)