I was preparing for the exam, and then I saw a few questions I did not understand regarding number of solutions in combinatorics. I would be glad if you could take a look and correct me if I'm wrong:

(questions are true\false):

1) The number of natural solutions of $x_1+x_2+x_3+x_4=14$ is equal to the number of positive whole number (integer) solutions of: $x_1+x_2+x_3+x_4=10$

2) The number of the positive even natural number solutions of $x_1+x_2+x_3+x_4=14$ is equal to the number of the positive odd natural number solutions of $x_1+x_2+x_3+x_4=14$

3) The number of integer solutions that are either $1$ or $-1$ to $x_1+x_2+x_3+x_4 \geq 0$ is 10.

4) If $|a|=3^3$ and $|b|=3$, then the number of functions from $a$ to $b$ is equal to the number of functions from $b$ to $a$.

What I did:

1) False. The number of the natural solutions of $x_1+x_2+x_3+x_4=14$ is $\binom {14+4-1}{3}$ and the number number of positive integer solutions of $x_1+x_2+x_3+x_4=10$ is $\binom {10+4-1}{3}$. It's false because the results are not the same.

2) False. The number of positive odd natural solutions of $x_1+x_2+x_3+x_4=14$ can be obtained if we divided the result by $2$ ($y_i=x_i/2$), i.e: $y_1+y_2+y_3+y_4=7$, so $\binom {7+4-1}{3}$. and for the even, we need that $x_i \geq 1$, so $y_i=\frac{x_i-1}{2}$, so $\binom {5+4-1}{3}$.

3) I don't know how to solve it, but I'll try. If we said the solution is only $1$, then there were no limits, and the result would be: $4!=24$. However, since it's more larger than $0$, we must have more $1$s than $-1$s, so we have to subtract the cases were we have more $-1$ than $1$, i.e: if at least $3$ variables are $-1$, so we subtract $2$ possibilities (either $3$ are $-1$ or $4$ are $-1$), and we get that the number of solutions is $24-2=22$ (If it was $3!-2$, it would make sense. I think I did it wrong).

4) False. If $|a|=3^3=27$ and $|b|=3$, then the number of functions from $a$ to $b$ is not the same as the number of functions from $b$ to $a$.

Please help me and correct me if I made any mistakes. Did my best to elaborate and explain what I've done. Thank you in advance!

  • $\begingroup$ I think you need to clarify what you mean by natural numbers (is $0$ natural?) and by whole numbers (since in $\# 3$ you appear to say that $-1$ is a whole number). $\endgroup$ – lulu Sep 3 '17 at 19:31
  • $\begingroup$ It is sometimes hard to follow your reasoning. for $\#2 $, for example, you are trying to count the "non-even" solutions, by which I assume you mean the solutions with four odd entries) but then you divide each term by $2$ which makes no sense unless they are even. $\endgroup$ – lulu Sep 3 '17 at 19:34
  • 1
    $\begingroup$ For $\#4$, your argument appears to be "this is false because it is false". Not good enough. $\endgroup$ – lulu Sep 3 '17 at 19:36
  • $\begingroup$ thank you for your comments. regarding 2: i did so because if i just divided by 2, i would get the uneven(for even i have to make sure it's larger than 1), so this is why i choose it to be as such. regarding 3: 1 is a whole number(integer as far as i know). is my calculation there was wrong? regarding 4: it seems obvious. was my decision wrong there? thank you again for your comments $\endgroup$ – BeginningMath Sep 3 '17 at 20:05
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    $\begingroup$ Please clarify whether $0$ is considered a natural number. $\endgroup$ – N. F. Taussig Sep 3 '17 at 20:08

1) FALSE $$ \eqalign{ & \quad \quad \left( {x_{\,1} + 1} \right) + \left( {x_{\,2} + 1} \right) + \left( {x_{\,3} + 1} \right) + \left( {x_{\,4} + 1} \right)\quad \left| {\;0 \le x_{\,k} \in Z} \right.\quad \Leftrightarrow \cr & \Leftrightarrow \quad x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} \quad \left| {\;1 \le x_{\,k} \in Z} \right. \cr} $$ It would be true the reverse (exchanging 10 and 14).

2) FALSE $$ \eqalign{ & {\rm N}{\rm .}\,{\rm sol}{\rm .}\,{\rm of}\;\left\{ {2x_{\,1} + 2x_{\,2} + 2x_{\,3} + 2x_{\,4} = 14\quad \Rightarrow \quad x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 7\quad \left| {\;1 \le x_{\,k} \in Z} \right.} \right\} < \cr & < {\rm N}{\rm .}\,{\rm sol}{\rm .}\,{\rm of}\;\left\{ {2x_{\,1} - 1 + 2x_{\,2} - 1 + 2x_{\,3} - 1 + 2x_{\,4} - 1 = 14\quad \Rightarrow \quad x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 9\quad \left| {\;\;1 \le x_{\,k} \in Z} \right.} \right\} \cr} $$


The number of $-1$'s can be $0$ or $1$ or $2$, corresponding to ${ 4 \choose 0}+{ 4 \choose 1}+{ 4 \choose 2}=11$ ways to place them.


Understanding $|a|$ as the number of elements of the set $a$, then the functions $f:a \to b$ are $|b|^{|a|}$ and those $f:b \to a$ are $|a|^{|b|}$

Thanks to N.F. Taussig for pointing out the correct interpretation of question (4), and other misunderstandings.

  • $\begingroup$ Your answer to the first question is incorrect. Notice that at most three of the summands can be equal to zero. $\endgroup$ – N. F. Taussig Sep 3 '17 at 20:55
  • $\begingroup$ @N.F.Taussig: sorry, you are right, I confused $\mathbb N$ with $\mathbb Z$ $\endgroup$ – G Cab Sep 3 '17 at 21:07
  • $\begingroup$ so 1 is right or wrong? i see the post was edited $\endgroup$ – BeginningMath Sep 3 '17 at 21:16
  • $\begingroup$ @BeginningMath: it was wrong the motivation (confusing N with Z). I amended now. But it remains false. $\endgroup$ – G Cab Sep 3 '17 at 21:22
  • $\begingroup$ The first question asks you to compare the number of nonnegative integer solutions to $x_1 + x_2 + x_3 + x_4 = 14$ and the number of positive integer solutions to $x_1 + x_2 + x_3 + x_4 = 10$. You seem to be answering a different question. Also, while statement 2 is false, your $x_i$s are positive integers for the even number case and nonnegative integers for the odd number case. The answer to the fourth question is also false. To answer it, you have to count the number of functions in each direction. Your answer to the third question is correct. $\endgroup$ – N. F. Taussig Sep 3 '17 at 21:26

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