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
!

SINGH SINGH, 12 years ago

Grade:

3 Answers

Ashwin Muralidharan IIT Madras

290 Points

12 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 x^r is ^n+r-1C_r {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).

Approved

SINGH SINGH

20 Points

12 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

12 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).