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Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ be a variable vector such that $\vec{r}.\hat{i},\vec{r}.\hat{j}$ and $\vec{r}.\hat{k}$ be positive integers.If $\vec{r}.\hat{j}\geq3$ and $\vec{r}.\vec{a}\leq12$,then the total number of possible $\vec{r}$ is equal to
$(A)\binom{10}{3} \hspace{1cm}(B)\binom{11}{3}\hspace{1cm}(C)\binom{13}{4}\hspace{1cm}(D)\binom{13}{9}$


As $\vec{r}.\hat{i},\vec{r}.\hat{j}$ and $\vec{r}.\hat{k}$ be positive integers,so $x>0,y>0,z>0$.Also $\vec{r}.\hat{j}\geq3$,so $y\geq3$
And $\vec{r}.\vec{a}\leq12$ means $x+y+z\leq 12$
Now i am stuck how to solve further.I am not able to apply stars and bars rule here.Answer given in the book is $\binom{10}{3}.$I do not know how.
Please help me.Thanks.

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  • $\begingroup$ Are you studying for some entrance exam? All your questions are multiple choice! :D $\endgroup$ – H. R. Nov 4 '15 at 12:41
  • $\begingroup$ We can Write it as $x+y+z+t = 12\;,$ Where $t>0$ and $x,y,z,t\in \mathbb{N}$. Now Using Star and bar Method , We can divide $12$ Stars into $4$ person such that each get at least $1$ star. Means we have $11$ space in between $12$ star and then we use $3$ seperatos in between $11$ places. This can be choose as $\displaystyle \binom{11}{3}$ $\endgroup$ – juantheron Nov 4 '15 at 12:44
  • $\begingroup$ Consider using \imath and \jmath to give $\imath$ and $\jmath$ respectively. That way, when you put hats on them you get $\hat\imath$ and $\hat\jmath$ instead of $\hat i$ and $\hat j$. $\endgroup$ – Fly by Night Nov 4 '15 at 14:12
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We have $x\ge1$ , $y\ge3$ and $z\ge1$. Then our inequation is equivalent to $a+b+c\le7$ where $a,b,c$ are non-negative integers.

Now if the sum $a+b+c$ is less than 7, then something added to it must equal 7. So we introduce a 'dummy' integer variable $d\ge0$ such that $$a+b+c+d=7$$

The number of solutions to that equation is clearly $\binom{10}{3}$

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