@bodhi_vriksha so do we have to remember this .....isn't there a proof for this?
There is...if you know your PnC well.
First let us examine for number of terms in (a+b+c)^n:
Effectively n powers have to be distributed into 3 groups (a or b or c) and groups can be empty (as there cound be terms like A^n where b and c have power 0!). This is equivalent to distributing n identical objects into 3 distinct groups and we can apply the theory of partitioning to figure out that this will be (n+2)C2 which as Team BV pointed out, will be (n+2)(n+1)/2
For example, CAT 08 had a question asking the number of terms in (a+b+c)^20 which would just be 22C2 or 22*21/2 = 231.
I will do a separate post for the coefficient, uska bhi PnC waala accha logic hai 😃 regards scrabbler
(1+x+x^5)^15As posted by @bodhi in one of his prev posts..No. of terms of a trinomial expanision is (n+1)(n+2)/2i.e 17*8 = 136.But there will be repetition of terms for x^5, x^10 and x^15.Hence, 136 - 3 = 133.PS : I'm not sure about my answer. Please do tag me with the OA and solution.!
Lot more repetition buddy....x^6 bhi repeat hoga na for example (1^13 *(x)^1 *(x^5)^1) and (1^9 *(x)^6 *(x^5)^0) and many other such.
A point to note: Minimum power is x^0. Max is (x^5)^15 = x^75. So there could be max 76 terms theoretically. Ab usmein se try to think what is not possible 😃
Shayad 70 aana chahiye, trying to solve orally 😃 Someone please do the hard work and confirm... @ChirpiBird 😁 regards scrabbler
@bodhi_vriksha so do we have to remember this .....isn't there a proof for this?
Basically, when we have something like (a+b+c+d)^10, take x=a+b and y=c+d.
Then expand (x+ y)^10 binomially. After that we need to do multiple binomial expansions on (a + b)^m and (c+d)^m , where m can take integer values.
Actually in our case, which just involves three variables, the coefficients are results of Pascal's tetrahedron as is the case in a binomial coefficients, which are results of Pascal's triangle.
In general, for a multinomial (x1+x2+...+xm)^n, any term of the expansion is given by nC(z1,z2,z3,..zm)*x1^z1*x2^z2*x3^z3*...*xm^zm, such that z1+z2+z3+...+zm = n
The coefficient is nC(z1,z2,z3,..zm)= n!/(z1!z2!...zm!)
The number of terms in the expansion of (x1+x2+...+xm)^n is (m+n-1)Cn or (m+n-1)C(m-1)
guys pls help me with this---->what is the coefficient of x^2*y^3*z^5 in (x+y+z)^10?
Statutory warning: Loooong post 😁
Let me start at a slight tangent.
In how many ways can 12 people be divided into three groups A, B, C of size 5, 4, 3?
Clearly this is 12C5*7C4*3C3 which if we expand turns out to be 12! / (5!*4!*3!). Now how come this beautifully simple formulas using the numbers above 5, 4, 3? Can't be a coincidence right 😃 So let's look more closely at it.
Let us make the 122 people stand in a line. Now take 12 post-it notes, write A on 5 of them, B on 4, and C on 3. Walk down the line and randomly stick a post-it on each person's forehead. They will get distributed into groups. But how many ways cvan it happen? The same as the number of words formed by arranging the letters AAAAABBBBCCC i.e. 12!/ (5!*4!*3!).
Thus dividing into groups which are not arranged internally means dividing by that group size ka factorial. In fact if we note carefully the basic formulae for nPr and nCr are special cases of the same general logic:
If we want to arrange r out of n people, the remaining (n-r) are a single group and are not being arranged so we divide the overall n! by (n-r)!.
In the case of selecting r out of n, we ware basically divding the n people into two groups (Selected and Rejected) neither group being internally arranged, so we divide by both r! and (n-r)!
Now (finally!! ) coming to the problem mentioned above i.e. what is the coefficient of x^2*y^3*z^5 in (x+y+z)^10?
When we expand (x+y+z)^10 = (x+y+z)(x+y+z)...(x+y+z)(x+y+z) 10 times, every term will be of order 10. So basically we have to choose which brackets will yield an x, which a y and which a z. If we want x^10, all brackets should yield x, so it is easy, only 1 way. But if we want say x^3*y^3*z^4? Easy enough, we need to allot 3 brackets to x, 3 to y and 4 to z which can be done in (you guessed it!) 10! / (3! 3! 4!) ways.
So for the question above the coefficient of x^2*y^3*z^5 in (x+y+z)^10 will be nothing but 10! / (2! 3! 5!)
In fact, when you solve problems such as whole number solutions of a+b+c+d+e=14, you are actually calculating the coefficient of x^14 in the expansion of (1+x+x^2+x^3.....)^5. This coefficient is the same as (14+5-1)C(5-1) or 18C14, which we guys also use as shortcuts.
@bodhi_vriksha@scrabbler how to find non existing terms . total terms will be 76 as pointed out above(1 + x + x^5)^15 - no of terms ?
Highest is 75.
Now next highest is 71 (x^5 14 times and x once) and then 70 (missing 3 terms 74 73 72)
Uske baad x^5 13 times and x twice gives 67, then 66 and 65 (missing 2 terms 69, 68)
Uske baad with x^5 12 times and x thrice we get 63 (missing 1 term 64)
Uske aage shayad sab milega....
So 6 terms not possible out of 76 giving 70 I feel...could be wrong! Double-check please.
Basically this is equivalent to solving for non-negative integers the following:
Given p + q + r = 15 how many values can q+5r take? OR also equivalent to solving something like:
If you take a test with 15 questions and a marking scheme of +5 for correct, +1 for no attempt and 0 for wrong, how many different scores are possible? regards scrabbler
What is the number of distinct triangles with integral valued sides and perimeter 14?
Let the three sides have length a, b, c such that a + b + c = 14 where a, b, c are positive integers.
But as a, b, c are three sides of a triangle, any integer values won't work. According to triangle inequality of sides (i.e. longest side is shorter than sum of other two side lengths), we get that longest side cannot be more than 6.
So if I give all the three sides 6 each then I have distributed 18 in them but perimeter was 14 only. That means I need to take out 4 from them. And if I take out a', b' and c' from the three sides respectively, then basically I am looking for unordered whole number solutions of the equation
a' + b' + c' = 4
which is C(6, 2) = 15.
Now this 15 is not the answer as it gives the number of ordered triplets. To find the answer, we need to un-order it.
See that there are some cases (of the form a,a,b) counted 3!/2! = 3 times because they had two numbers same - i.e. (004), (112), (220) ...3 cases.
And some cases (of the form abc) counted 3! = 6 times because they had all three numbers different - let it be x.
So we can write that 15 = 6x + 3(3) i.e. x = 1 and the required number of triangles is = x + 3 = 4. :)
ALTERNATELY: You could have simply calculated the cases by fixing the longest side length as follows - 662 653 644 554 i.e. 4 triangles..:)
Ext. Qs. What is the number of distinct triangles with integral valued sides and perimeter 2014?
Lot more repetition buddy....x^6 bhi repeat hoga na for example (1^13 *(x)^1 *(x^5)^1) and (1^9 *(x)^6 *(x^5)^0) and many other such.A point to note: Minimum power is x^0. Max is (x^5)^15 = x^75. So there could be max 76 terms theoretically. Ab usmein se try to think what is not possible Shayad 70 aana chahiye, trying to solve orally Someone please do the hard work and confirm... @ChirpiBirdregardsscrabbler
Bro how will this term repeat i mean (x^5)^0 = 1 right?
No a+b + c = 15 then b + 5c = how many possible values? is what you must solve.regardsscrabbler
c can take 0-15 so totally 15 values b can take 0-15 so totally 15 values in both the above method the distribution is mutually exclusive plus 1 when any of the two is zero. now consider when b and c tootally makke up 15 (1,14)=71 (2,13)=67 (3,12)63 (4,11)=59
(5,10)=55 (6,9)=51 (7,8)=47 Now when both b,c are equal (1,1)=6 but this value is got so omit it (2,2)=12 omit this too (3,3)=18 (4,4)=24 (5,5)=29 (6,6)=26 (7,7)=42
So i guess totally there should be 31+7+5=43 Is my approach right?
😁 
you're right.! lemme review it..
a+b + c = 15 then b + 5c = how many possible values? is what you must solve.