there are 3 friends a,b,c they all have some 1rupee coins. a has 6 and b and c have 7 and 8 cons respectively. they want to donate rs 10 to a trust . in how many ways they can do it

pls explain using multinomial theorem and pls tell what is multinomial theorem.

Let x,y,z be number of coins denoted by a,b,c respectively
So we are given
x+y+z=10
x can range from 0-6 since a can choose 0 to 6 coins
similarly 
y can range from 0-7
z can range from 0-8
The number of ways in which they can donate money
=(x0+x1+x2+x3+x4+x5+x6)(x0+x1+x2+x3+x4+x5+x6+x7)(x0+x1+x2+x3+x4+x5+x6+x7+x8)=(x7-1)(x-1)(x8-1)(x-1)(x9-1)(x-1)=(x7-1)(x8-1)(x9-1)(x-1)-3 =(1-x7)(1-x8)(1-x9)(1-x)-3 =(1-x7)(1-x8)(1-x9)(1+3c1x+4c2x2+.........+12c10x10+...) Now since they have to donate 10 rupees so we will find coefficient of x10 in above expressio Coefficient of x10 =(12c10-5c3-4c2-3c1) =66-10-6-3 =47 So total we have 47 ways
 

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x+y+z = 10

x,y,z are the number of coins deonated by a,b,c respectively.

x can be any value from 0 to 6 , because 'a' can donate 0 coins to 6 coins .

Similarly, y and z are from '0 to 7' and '0 to 8' respectively.

So, the coefficient of x10 in the expression (given below) are the no. of ways in which they can donate 10 coins.

(x0+x1+x2+ x3+x4+x5+x6)(x0+x1+ x2+x3+x4+ x5+x6+x7)(x0+x1+x2+ x3+x4+x5+ x6+x7+x8)

=[1(x7-1)/(x-1)] .[1(x8-1)/(x-1)] .[1(x9-)/(x-1)]

=(1-x7)(1-x8)(1-x9)(1-x)-3

=(1-x7)(1-x8)(1-x9) (1+ 3C1x+ 4C2x2+ 5C3x3+ 6C4x4+.......+ 12C10x10+...........) [because, (1-x)-n=1+ nC1x + n+1C2x2+ n+2C3x+ n+3C4x+ n+4C5+........]

= (12C10 -5C3 -4C2 -3C1)x10+...... other terms.

= 66 -10 -6 -3

= 47 ways

I hope, it helps you.

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Very intelligent answer.

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