Kenneth Rosen Edition 6th Exercise 5.5 Question 15 (Page No. 380)

Question

How many solutions are there to the equation
x1 + x2 + x3 + x4 + x5 = 21,
where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that: 0$\leq$ x1$\leq$10 ?

Comments

Same as this:

https://gateoverflow.in/132586/discrete-maths-counting-number-of-solutions-to-the-equation

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Best explanation! 

https://www.slader.com/discussion/question/how-many-solutions-are-there-to-the-equation-x_1-x_2-x_3-x_4-x_5-21/ 

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@asqwer Please do not spam with links

Answer 1

x1 + x2 + x3 + x4 + x5 = 21, where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that: 0≤ x1 ≤10 and x2,x3,x4,x5 can range in between 0 to 21.

Now we have to choose the coefficient of x²¹ in following expression

(1+ x + x² +x³.....+ x¹⁰) (1+ x + x² +x³.......+X²¹)(1+ x + x² +x³.....+ x²¹)(1+ x + x² +x³.....+ x²¹)(1+ x + x²+x³.....+ x²¹)

also we can extend term of x2,x3,x4,x5 from x²¹ upto infinty because it didn't  change the value of x²¹ coefficient . 

(1+ x + x² +x³.....+ x¹⁰) (1+ x + x² +x³.......∞)(1+ x + x² +x³.....+ ∞)(1+ x + x² +x³.....+ ∞)(1+ x + x²+x³.....+ ∞)

$\left ( \frac{1-x^{11}}{1-x} \right )\left ( \frac{1}{(1-x)^{4}} \right )$

$\left ( 1-x^{11} \right )\left ( \frac{1}{(1-x)^{5}} \right )$

now we have to find only coefficient of x²¹ for which we consider coefficient of x²¹ and x¹⁰ in $\left ( 1-x \right )^{-5}$

general term of binomial coefficient is $\binom{n+r-1}{r}$

$\binom{5+21-1}{21}-\binom{5+10-1}{10}$=12650-1001=11649

Comments

great explaination sir...thanx a lot :) :)

Answer 2

x1 + x2 + x3 + x4 + x5 = 21, where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that: 0≤ x1 ≤10 .

Which can be solved by formula :  $\binom{25}{4} - \binom{14}{4}$

Total possibilites without upper Bound : $\binom{25}{4}$

possibilites with upper Bound: $\binom{14}{4}$

Answer will be 11,649.

Comments

condition will be like this:

xi>=0, So 25c4

we will remove all solutions where x1>=11 and xi>=0

so x1 + x2 +x3 +x4 +x5 =21

   make x1>=0  by removing 11 from 21

x1 +x2 + x3 +x4 +x5 =10

(10 + 5-1 )c(5-1)

check this

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Thank you for pointing . Agreed. :)

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i was solving for x1+x2+x3+x4+x5=11, silly mistake... thanx for helping :)

Answer 3

x1 + x2 + x3 + x4 + x5 = 21

0≤x1 ≤3  ,1≤ x2<4  and x3≥15

case 1: x3≥15 given to make condition <x1, x2, x3, x4, x5> ≥ 0 remove 15 from 21.

so new equation becomes x1 + x2 + x3 + x4 + x5 = 21-15 =6

case 2: 1≤ x2<4=1≤ x2≤3 given so make it 0≤ x2 ≤2   for that remove 1 from 6. 

so new equation becomes x1 + x2 + x3 + x4 + x5 =6-1=5

find all solutions where <x1, x2, x3, x4, x5> ≥ 0 lets say it solution A.

and remove the solution where 0≤ x2 ≤2 that is we will remove all solutions from A where x2>2 and all other variables are xi>=0 let's say it solution B.

 0≤x1 ≤3 that is we will remove all solution from A where x1>3 and all other variables are xi>=0 let's say it solution C. 

^{A= (5+5-1)}C ^_(5-1)     

   = 126

C= 5C4 =5 solutions 

B=6C4 =15 solutions 

B and C =0 solutions

Answer = A -  (B+C - B and C) 

               =126-20

               =106

Comments

Can you explain how u got case B

Answer 4

x1 + x2 + x3 + x4 + x5 = 21, where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that: 0≤ x1 ≤10 ,0≤ x2 ≤10 ,0≤ x3 ≤10 ,0≤ x4 ≤10 ,0≤ x5 ≤10 ,

Solved by Generating Function :

(1+ x + x² +x³.....+ x¹⁰) (1+ x + x² +x³.....+ x¹⁰)(1+ x + x² +x³.....+ x¹⁰)(1+ x + x² +x³.....+ x¹⁰)(1+ x + x² +x³.....+ x¹⁰) = x²¹.

((1+ x + x² +x³.....+ x¹⁰))⁵ = x²¹.

$\left ( \frac{1- x^{11}}{1-x} \right )^{5}$ = x²¹.

$\left ( 1- x^{11} \right )^{5}\left ( \frac{1}{1-x} \right )^{5}$ =  x²¹.

$\left ( 1- x^{11} \right )^{5}\sum_{r=0}^{n}\binom{n+r-1}{r} x^{r} = x^{21}.$

$\left ( 1- x^{11} \right )^{5}\sum_{r=0}^{n}\binom{5+r-1}{r} x^{r} = x^{21}.$

$\left ( 1- x^{11} \right )^{5}\sum_{r=0}^{n}\binom{r+4}{r} x^{r} = x^{21}.$

Now we have to find term of x whose cofficient is 21.

$\left (\binom{5}{0} x^{0} - \binom{5}{1}x^{11} + \binom{5}{2}x^{22} - \binom{5}{3}x^{33}+ \binom{5}{4}x^{44} - \binom{5}{5}x^{55} \right )$  $\times \sum_{r=0}^{n}\binom{r+4}{r} x^{r} = x^{21}.$

$\left ( _{0}^{5}\textrm{C} \times _{21}^{25}\textrm{C}x^{21} \right ) - \left ( _{1}^{5}\textrm{C} \times _{10}^{14}\textrm{C}x^{21} \right ) = x^{21}$

$x^{21}\left ( 1 \times \left ( 25\times 23\times 22 \right ) \right ) - \left ( 5 \times\left ( 7\times 13\times 11 \right ) \right ) = x^{21}$

= 7645

Answer 5

$\begin{align*}
&\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 21 \quad \left ( 0\leq x_i \leq 10 \right ) \\ \\
&\Rightarrow \text{No of integral solution of the above equation is } \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left ( 1+x+x^2+ \dots + x^{10} \right )^5 \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left [ \frac{1-x^{11}}{1-x} \right ]^5 \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left [ 1-x^{11} \right ]^5 \cdot \left [ \frac{1}{\left ( 1-x \right )^5} \right ] \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left [ \sum_{m=0}^{5}\binom{5}{m}\left ( -1 \right )^m \cdot x^{11\cdot m} \right ] \cdot \left [ \sum_{r=0}^{\infty}\binom{5+r-1}{r}\cdot x^r \right ] \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \begin{cases}
&+1 \cdot \binom{25}{21} \qquad m = 0,r = 21 \\ \\
&-5 \cdot \binom{14}{10} \qquad m = 1,r = 10 \\
\end{cases} \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} = +1 \cdot \binom{25}{21} -5 \cdot \binom{14}{10} = {\bf \color{red}{7645}} \\
\end{align*}$

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Comments

but constraint is for x1 that it lies btw 0 and 10.why r u taking it for all??