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Explain each one of the following:

a ) In how many ways can we put 31 people in 3 rooms such that each room has an odd number of people ?  

b ) Coefficient of $x^4$ in the expansion  $(1+ x + x^2 + x^3)^{11}$ using generating functions.

c)Find out number of solutions $x_1+x_2+x_3 = 20 , 2<x_1<6 , 6<x_2<10 , 0<x_3<5$
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4 Answers

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Quetion B. 

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part (a)

we need to find out the coefficient of $x^{28}$ in the last expression.

$$\begin{align*} &=\left [ x^{31} \right ] \left ( x^{1}+x^{3}+x^{5}+x^{7}....\infty \right )^{3} \\ &=\left [ x^{31} \right ] \left [ x.\left ( 1+(x^2)^1+(x^2)^3+...\infty \right ) \right ]^{3} \\ &=\left [ x^{31} \right ]\left [ \frac{x}{1-x^2} \right ]^{3}\\ &=\left [ x^{31} \right ]\left [ \frac{x^{3}}{\left ( 1-x \right )^{3}.\left ( 1+x \right )^{3}} \right ] \\ &=\left [ x^{28} \right ]\left [ \sum_{r=0}^{\infty}\binom{2+r}{r}.x^r \right ] \left [ \sum_{r=0}^{\infty}\binom{2+r}{r}.(-1)^r.x^r \right ] \end{align*}$$

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1 votes

A. 120

Room1: 2a+1

Room2: 2b+1

Room3: 2c+1

(2a+1) + (2b+1) + (2c+1) = 31

a+b+c = 14. 

whose solution is C(14+3-1,3-1) = C(16,2) = 120

B. Don't know

C. 0

number of solutions x1+x2+x3=20   (Solve it using Multinomial Theorem)

is coefficient of x^20 in (x^3+x^4+x^5)^2  *  (x+x^2+..x^4)

i.e coefficient of x^13 in (1+x+x^2)^2  *   (1+x+...x^3)

Since highest power in the above equation is x^7. Therefore coefficient of x^13  will be zero.

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1 votes

For B using generating functions.

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