sir, how you take a,b,c,d as different groups at one time different to calculate no of a^11,b^3,c^6,d^10 and at another time whole as same to calculate total possible 30 of a,b,c,d?
Vinit, I haven’t taken them as the same. I have just said that the total will be 30! The thing is when we multiply 30 brackets as (a+b+c+d)(a+b+c+d)… each term will be created by picking one element from each of the 4 brackets. This will be either a or b or c or d, so overall there definitely will be 30 variables picked (one per bracket) but the distribution among those variables might vary…
Refresh your memory with this https://crackthecat.wordpress.com/2013/03/11/the-binomial-theorem/ and you will be able to see the parallels I think 🙂
Thank u so much sir, now i understood…!!
Sir I read no. of terms in expression (1+x+x^2)^12 is 13 since power of x will vary from zero to 13 . Then, how many terms will be there in expression (1+x+x^3)^12 ?
Sorry Sir Mistake in above post I read no. of terms in expression (1+x+x^2)^6 is 13 since power of x will vary from zero to 13 . Then, how many terms will be there in expression (1+x+x^3)^6?
Pranita, it seems basically the same as saying what amounts can you make with coins of 0, 1 and 3 rs (6 of each type). So I would say all values from 1 to 18 possible except 17…. so 18 cases I guess?
Sir I dont know the answer for this question.I saw similar question so thought wat would be the answer for this.
Hi J! I understood the approach for finding the number of terms and co-efficient of a particular term through PnC. But kindly explain if that approach be extended to same variable with different powers e.g. (x-x^(-2)+x^3)^3? I tried to find a pattern for a couple of such cases but in vain. So is expansion into individual terms the only way out in these cases? And I read your approach in the above comment of forming different amounts with 3 denominations…but that too, in my opinion, will require quite a bit of iterations. So kindly give some insight in these scenarios and if possible a quicker approach.
That becomes a little trickier, especially in the cases where some intermediate powers are missing (1 + x + x^4) type things. I haven’t really focused much on that ever as only very simple variants have come in CAT-type exams till date (which can be quickly done by checking for a small value and observing pattern) so I don’t have a generic approach; will think on it and if I can come up with an efficient approach will share the same 🙂
Thanks a lot J. Then can we can safely assume that CAT is only going to cover the types of (a+b+c+d+…n)^r where no two variables will be same?
And in the above comment you mentioned about the intermediate powers being absent, then kindly give a approach for a more conventional type but of same variable type e.g (x+x^2+x^3)^3 or (x^(-1)+1+x)^3 or simply (x+x^2)^3.
I am sure if the variables are all different, individual powers of them of any magnitude and nature (+ve/-ve) and can be dealt with your approach pretty effectively(I’ve tried out for quite a lot and it worked like a charm). Only the same variable ones are posing to be a tough nut to crack. It will be really helpful if you can share your way of approach if faced with such a type, be it the simplest one. Thank you.
In the expansion of (a+b+c+d)^20 suppose we want to find the coeff of a^5*b^5*c^5*d^5 then should it be 20!/5!5!5!5! or 20!/5!5!5!5!4!
20!/5!5!5!5! as we have 4 distinct groups a b c and d to distribute the powers amongst.
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