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Hi guys this is my first time using this website and I thought it would be useful as I am currently in 7th grade doing Maths Olympiad and studying permutation and combination here is my question ---->

Mrs Reed forgets the combination number to her safe. The combination has 4 numbers, each a 1 digit number. The 4 digits add up to 10 and there is no zero. How many times at the most does Mrs reed have to try to open her safe?

Thanks and pls note that I am currently in Grade 7 so pls explain it so I am able to "understand"

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    I have tried writing it out in numbers but I know there is definitely more efficient ways to do it – 2017-02-03
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    Take a look at this https://en.m.wikipedia.org/wiki/Stars_and_bars_(combinatorics)#Statements_of_theorems – 2017-02-03
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    Come on: a first time visitor here of school age does not deserve to receive -6 in feedback for what is a perfectly fair (but duplicated) question. – 2018-05-18

2 Answers 2

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Ok, so the combination consists of four digits $(a,b,c,d)$ where each of them is greater than $0$ and $a+b+c+d=10$. Let's assume you are allowed to repeat digits (though you do not say this) and count by brute force.

First note that none of the digits can exceed $7$ (as $8+1+1+1>10$). We'll work from the maximum of the digits.

Case I: Max is $7$. Then the password is $\{7,1,1,1\}$ in some order. There are $4$ places to put the $7$ so we get $\boxed {4}$ cases.

Case II: Max is $6$. Then the password is $\{6,2,1,1\}$ in some order. Now there are $4$ places to put the $6$ and then $3$ places to put the $2$, so $\boxed {12}$ cases.

Case III: Max is $5$. Then we can have $\{5,3,1,1\}$ or $\{5,2,2,1\}$ Each of these gives us $12$ cases so $\boxed {24}$.

Case IV: Max is $4$ We can have $\{4,4,1,1\}$ ($\binom 42=6$ cases) or $\{4,3,2,1\}$ ($4!=24$ cases) or $\{4,2,2,2\}$ ($4$ cases) so $\boxed {34}$.

Case V: Max is $3$. We can have $\{3,3,3,1\}$ ($4$ cases) or $\{3,3,2,2\}$10 ($6$ cases) so $\boxed {10}$

Of course the maximum has to be at least $3$ (as $2+2+2+2<10$) so (barring error) these are all the cases.

Thus the total is $$4+12+24+34+10=\boxed {84}$$

Note: The brute force method is easy to understand but it is error prone. To approach the problem more systematically, you can read about Stars and Bars. From that method we see that the answer is $$\binom {9}3=\boxed {84}$$

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    I checked the answer its 80 unfortunately. – 2017-02-03
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    I think your answer key is wrong. – 2017-02-03
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    That surprises me. First of all, the Stars and Bars method is hardly error prone and it matches the result obtained by counting. Secondly, the most likely error in the counting method is that you skip a case...so I'd be less surprised if I undercounted. Perhaps you left off a condition? – 2017-02-03
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    I am confident its not because this problem showed up on many books and all answers are the same. Im confused. – 2017-02-03
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    Searching online turned up [this old question](http://math.stackexchange.com/questions/1052988/mrs-reed-safe-combination) and I note that the same debate occurred there (people here get $84$ where some reference is said to give $80$). – 2017-02-03
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    @VHSenpai $84$ is correct. Read here too: http://math.stackexchange.com/questions/1052988/mrs-reed-safe-combination – 2017-02-03
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See this is what they showed in the answer key ---->

Possible combinations --- Numbers of ways to arrange

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*1117 4 *1126 12 *1135 12 *1144 6 *1225 12 *1234 24 *2224 4 *2233 6

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    They appear to have skipped the four versions of $\{3,3,3,1\}$. – 2017-02-03
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    Like I say, the problem with the brute force method is that it is easy to skip a case or two. Here, there really aren't that many cases so I think it's possible to be quite confident....but if we had a ten digit combination that added to $50$ then errors would be a lot harder to avoid. Stars and Bars is an important method to learn! – 2017-02-03
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    OMG thank you so much I realize now okay thanks guys that was the last problem in my Maths Olympiad book. – 2017-02-03