Permutation and combination doubt

Question

In how many ways 3 friends ram ,rajat and rupesh having 6 one rupee coins, 7 one rupee coins ,8 one rupee coins respectively donate 10 rupees collectively.

a)40

b)45

c)47

d)50

Comments

47??

Answer 1

The given problem can be formulated as :

x1 + x2 + x3 = 10 subjected to constraints :

0 <= x1 <= 6 ,

0 <= x2 <= 7

0 <= x3 <= 8

So solution to the given problem is nothing but finding coefficient of x¹⁰  in (1 + x ..+ x⁶) . (1 + x .. + x⁷) . ( 1 + x ....+ x⁸)

                                                         ==>   coefficient of x¹⁰  in  (1 - x⁷) . (1 - x⁸) . (1 - x⁹ ) .( 1 - x ) ^-3

Now out of the terms of (1 - x⁷) . (1 - x⁸) . (1 - x⁹ ) , we extract only those terms which contribute to x¹⁰ ..

So

(1 - x⁷ ) . (1 - x⁸ - x⁹ + x¹⁷ )    =  1  -  x⁸  - x⁹  -  x⁷  [ Only those terms which contribute to x¹⁰ are mentioned ]..

Now for each of the 4 terms we find the corresponding coefficient in ( 1 - x ) ^-3  and then finally sum up to find the answer.

So due to the term 1 , we have :   ^3-1+10C₁₀   =   66

      due to the term 2 , we have :  ^3-1+2C₂  =  6

        due to the term 2 , we have :  ^3-1+1C₁  =  3

        due to the term 2 , we have :  ^3-1+3C₃  = 10

Hence , no of ways  =   66  -  6 - 3 - 10 

                              =   47

Hence 47 should be the correct answer.

Comments

@ Habibkhan sir

Can u plzz provide some guidance in solving question related to finding number of conflict serializable schedule...

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Plz post separate question on ur doubt of conflict serialisability , then I will be able to answer..By the way I m not a sir..I m a learner..:)

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https://gateoverflow.in/148707/no-of-conflict-serializable-schedule-doubt

This is the link

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Sorry to bother u again

So due to the term 1 , we have :   ^3-1+10C₁₀   =   66

      due to the term 2 , we have :  ^3-1+2C₂  =  6

        due to the term 2 , we have :  ^3-1+1C₁  =  3

        due to the term 2 , we have :  ^3-1+3C₃  = 10

How did u write this...can u Plzz help

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Coefficient of x^r  in (1 - x)^-n  is given by : ^n-1+rC_r

So say I have term 1 which is '1' so we need coefficient of x¹⁰ in (1 - x)^-3 ..

Likewise for x⁷ , in order to get x¹⁰ , we need x³ , hence we find coefficient of x³ in (1 - x)^-3 and so on..

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Thankyou soooooo much....sir....

U r very  kind and helpful  ...:)

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is there any shortcut  method it is time consuming ..too long method but thanks sir

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So due to the term 1 , we have :   ^3-1+10C₁₀   =   66

      due to the term 2 , we have :  ^3-1+2C₂  =  6

        due to the term 2 , we have :  ^3-1+1C₁  =  3

        due to the term 2 , we have :  ^3-1+3C₃  = 10

*please explain ^3-1+10C₁₀   =   66(how    ¹⁰C₁₀   )

due to the term 2 , we have :  ^3-1+1C₁  =  3 (how ¹C₁)

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So due to the term 1 , we have :   ^3-1+10C₁₀   =   66

      due to the term 2 , we have :  ^3-1+2C₂  =  6

        due to the term 2 , we have :  ^3-1+1C₁  =  3

        due to the term 2 , we have :  ^3-1+3C₃  = 10

*please explain ^3-1+10C₁₀   =   66(how    ¹⁰C₁₀   )

due to the term 2 , we have :  ^3-1+1C₁  =  3 (how ¹C₁

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3-1+10C10   =   10+3-1C10 = 12C 10 = 66