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Hi. I was going through your free study material about the multinomial theorem. please consider the following steps coefficient of x^10 in (1-x^7)(1-x^8)(1-x^9)(1-x)-3 (I got it till here) Now I have no idea whats going on in the next step .. that is = coefficient of x10 in (1-x7-x8-x9)(1+3C1x + 4C2x2+ 5C3x3+...+12C10x10) (ignoring powers higher than 10) = 12C2 - 5C3 - 4C2 - 3C1 = 66 - 10 - 6 - 3 = 47. Please help me out ASAP !

Hi.
I was going through your free study material about the multinomial theorem.

please consider the following steps

coefficient of x^10 in (1-x^7)(1-x^8)(1-x^9)(1-x)-3

(I got it till here) Now I have no idea whats going on in the next step .. that is

= coefficient of x10 in (1-x7-x8-x9)(1+3C1x + 4C2x2+ 5C3x3+...+12C10x10)

(ignoring powers higher than 10)

= 12C2 - 5C3 - 4C2 - 3C1

= 66 - 10 - 6 - 3 = 47.


Please help me out ASAP
!

Grade:

3 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Singh,

 

Co-eff of x10 in (1-x7)(1-x8)(1-x9)(1-x)-3 is what is required.

Multiplying only the first three terms you will get (1-x7-x8-x9+x15+x16+x17-x24)*(1-x)-3.

So for the co-eff of x10, only those powers of x which are less than or equal to 10 will make a contribution.

And (1-x)-n, the co-eff of xr is n+r-1Cr {So that is expanded again}

Now, to get co-eff of x10, say you take 1 from the first bracket, then you will take co-eff of x10 from the second bracket. Similarly to get co-eff of x10, if you take x7 from the first bracket, then you will take co-eff of x3 from the second bracket.

 

And hence the co-efficients are obtained.

 

All the best.

Regards,

Ashwin (IIT Madras).

SINGH SINGH
20 Points
9 years ago

Thankyou so much !that helped me clear the doubt !

Sir, I understood that how  coeff. are obtained but like ,

I have a problem

In a box there are 10 balls ,4 red,3 black,2 white and one yellow !in how many ways can you select 4 balls out of these 10 balls'??

 

solution ::-

let x1,x2,x3,x4 be the no. of red ,black ,white and yellow balls respectively .

no. of ways to select the 4 balls= x1+x2+x3+x4=4

coeff of x^4 in (1-x^5)(1-x^4)(1-x^3)((1-x^2)(1-x)^-4

NOW COMES MY DOUBT , 

My book says that the next step is ,

coeff of x^4 in (1-x^-4)- coeff of x^2 in(1-x^-4)-coeff of x^1 in(1-x^-4)-coeff of x^0 in (1-x^-4)

 

 

YOU SAID THAT COEFF. TERMS WHOSE POWER IS GREATER THAN RHS SHOULD BE REJECTED ! 

NOW THATS QUITE TRUE BUT WHY HAVE WE NOT CONSIDERED COEFF OF X^3 ?  ? ? ? ? ?

 

I HAVE'NT DONE THE CHAP OF BINOMIAL THEOREM SO THE DOUBT MIGHT SEEM SILLY BUT PLEASE HELP IF U CAN !!! :)


 

 

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Singh,

 

First of all, please don't address us as Sir. You can just address by the name.

 

Coming to your doubt:

 

Yes, the terms with powers >4, are neglected from the terms in the product of: (1-x^5)(1-x^4)(1-x^3)(1-x^2).

Plese note that the x^3 term is not neglected. It is because of the x^3 term, you are subtracting the co-eff of x^1.

 

If you evaluate this product and write only those terms whose power is less than or equal to 4,

Then those terms would be 1-x^2-x^3-x^4.....

 

And hence to get the co-eff, you would need co-eff of x^4 in (1-x)^-4 (because of the constant term "1"), minus co-eff of x^2 in (1-x)^-4 (this is because of the x^2 term in that product), minus co-eff of x^1 in (1-x)^-4 (this is because of the x^3 term) and finally the subtraction of constant term (because the product has x^4).

 

The product no-where has x^1 to take co-eff of x^3 from (1-x)^-4.

 

And I hope that clears your doubt. In case of any more doubts scrap me directly, and I can respond to your querry faster.

 

All the best.

Regards,

Ashwin (IIT Madras).

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