How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 21$,

Question

How many solutions are there to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 21$$,

such that

all of $$x_{i}$$ where $$i=1,2,3,4,5$$ are non negative

and

$$0\leq x_1 \leq 3$$

$$1\leq x_2 \lt 4$$

and

$$x_3 \geq 15$$

Attempt

first used $$0\leq x_1 \leq 3$$ and

$$1\leq x_2 \lt 4$$ and then find the number of solutions

violating $$0\leq x_1 \leq 3$$ $$\\,\,$$and$$\,\,1\leq x_2 \lt 4$$

will give $$x_1 \gt 3\\,\,$$and$$\,\, x_2 \gt 3$$

Now number of solutions are there to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 21$$,

=$$\binom{21+5-1}{21}=12,650$$

Now number of solutions are there to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 21$$, such that $$0\leq x_1 \leq 3$$ and $$1\leq x_2 \lt 4$$

=$$12650-$$number of solutions are there to the equation $$x_1 \gt 3\\,\,$$and$$\,\, x_2 \gt 3$$

$$-----------------------------------------------$$

solving number of equation for $$x_1 \gt 3\\,\,$$and$$\,\, x_2 \gt 3$$

let $$x_1=x_1^{'}+3$$

$$x_2=x_2^{'}+3$$

our equation becomes

$$x_1^{'}+3+x_2^{'}+3+x_3+x_4+x_5=21$$

$$\Rightarrow x_1^{'}+x_2^{'}+x_3+x_4+x_5=15$$

$$\therefore$$ number of equation=

$$\binom{15+5-1}{15}=3876$$

Now,

Now number of solutions are there to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 21$$, such that $$0\leq x_1 \leq 3$$ and $$1\leq x_2 \lt 4$$

=$$12650-3876=8,774$$

now among $$8,774$$ we have to find $$x3 \geq 15$$

for $$x_3 \geq 15$$,

let $$x_3 =x_3 ^{'}+15$$

our equation becomes

$$x_1 + x_2 + x_3^{'} +15+ x_4 + x_5 = 21$$,

$$x_1 + x_2 + x_3 + x_4 + x_5 = 6$$

$$\therefore$$ number of equation =$$\binom{6+5-1}{6}=210$$

so final answer$$=8,774-210=8,564$$

But the answer is $$106$$

Where am i wrong??

Please correct me or else give me the correct way

Thanks!

,

• Any constraints on $x_4$ and $x_5$? non-negative integers? May 16 '17 at 10:50
• no ! no any constraint on $x4$ and $x5$ May 16 '17 at 10:52
• yes all of $x_{i}$ where $i=1,2,3,4,5$ are non negative May 16 '17 at 10:52
• math.stackexchange.com/questions/397127/… May 16 '17 at 11:03
• @DaríoA.Gutiérrez link is of no use ,as i am asking for help "to give correction in my approach" May 16 '17 at 11:05

Let $S_{a, b, c, d, e}$ be the number of solutions with $x_1\ge a$, $x_2 \ge b$, $x_3\ge c$, $x_4\ge d$ and $x_5\ge e$.

The equation can be written as

$$(x_1-a)+(x_2-b)+(x_3-c)+(x_4-d)+(x_5-e)=21-a-b-c-d-e$$

So we have

$$S_{a, b, c, d, e} =\binom{21-a-b-c-d-e+4}{4}$$

if $a+b+c+d +e\le21$ and is $0$ otherwise.

The answer to this question is

$$S_{0, 1,15,0,0}-S_{4,1,15,0,0}-S_{0,4,15,0,0}+S_{4,4,15,0,0}=\binom{9}{4}-\binom{5}{4}-\binom{6}{4}+0=106$$

• got it @CY Kwong sir !thanks ! May 16 '17 at 12:25

see this answer for more explanations on the method

The answer is the coefficient of $x^{21}$ in

$$(1+x+x^2+x^3)(x+x^2+x^3)(1+x+x^2+\ldots)^3$$ which is the same as the coefficient of $x^{21}$ in

$$x(1-x^4)^2(1-x)^{-5}= x(1-2x^4 +x^8)\sum_{k=0}^\infty \binom{4+k}{k}x^k$$

which can be read off as

$$\binom{24}{20} -2\binom{20}{16} + \binom{17}{12}$$

• I wouldn't agree. If $1 \le x_{2} \le 4$, then why did you write $x+x^2+x^3$ and not $x+x^2+x^3+x^4$ ? Dec 13 '17 at 20:36
• @Karagum: It says $x_2\lt4$. Apr 15 '20 at 11:02
• There are two errors here. The factor $x^{15}$ from $x_3\ge15$ is missing, and one of the factors $1-x^4$ should be $1-x^3$ (since $x+x^2+x^3=x(1-x^3)(1-x)^{-1}$). Apr 15 '20 at 11:08