How do I solve binomial theorem questions for JEE?
Now, do they really? I don't find binomial theorem questions, especially of JEE within the scope of being difficult or unsolvable. There is some basic layers of knowledge, concept, that you are missing in your approach. So, I would recommend you go back to basics and come back again. I personally feel, a sample problem would be better for me to answer.However, I might take a guess, do these binomial problems have anything related to summation, if yes, all you have to do is convert it into an integral. That will just do the trick. Good luck!
JEE Advanced Doubts: How can I solve question number 13?
1st Of all, i'm not IITian or another -ian, i'm a IIT-JEE aspirant 2015, Quite-well prepared! :-) So you can trust on Me!Solution:As in this question already mentioned that,[math] n\ge\frac{k(k+1)}{2} [/math]and [math]x_1\ge 1[/math],[math]x_2\ge 2[/math] And so on..So, assume each [math] x_i\geq i+r_i,(i= 1,2,3...k) [/math]and r any integer!Then we get,[math] \sum_{i=1}^{k}x_i ={\sum_{i=0}^{k}r_i}+\frac{k(k+1)}{2} =n[/math]then, [math] {\sum_{i=0}^{k}r_i}=n-\frac{k(k+1)}{2} [/math] then the problem is obvious!You need to Fill [math] k[/math] box with [math] n-\frac{k(k+1)}{2} [/math] Balls..then Solution is [math]\binom{n-\frac{k(k+1)}{2} +k-1}{k-1}[/math]Problem in Understanding?Comment below!if my answer somewhere Helps you Don't be Silly for upvoting It!:-) you are Such a kind Person! Thank you!
Please help me with extremely difficult binomial expansion question?
nC1 x a x b^(n-1) = 12 -------------1) nC2 x a² x b^(n-2) = 60 -------------2) nC3 x a³ x b^(n-3) = 160 ------------3) eqn 2) / eqn 1) [n! / {2! * (n-2)!}] * a² * b^(n-2) ....60 --------------------------------------... = ------ [n! / {1! * (n-1)!}] * a * b^(n-1) .....12 [1 * (n-1)*(n-2)! ] *a * b^(n-2) ....60 --------------------------------------... = ------ ; we know (n-1)! = (n-1)*(n-2)! and b^(n-1) = b*b^(n-2) [2 * (n-2)! ] * b*b^(n-2) ............12 (n - 1) a ------------- = 5 2*b => a/b = 10/(n - 1) ----------4) Similarly eqn 3) / eqn 2) [n! / {3! * (n-3)!}] * a³ * b^(n-3) ....160 --------------------------------------... = ------ [n! / {2! * (n-2)!}] * a² * b^(n-2) .....60 [2! * (n-2)*(n-3)! ] *a * b^(n-3) ....8 --------------------------------------... = ------ ; we know (n-2)! = (n-2)*(n-3)! and b^(n-2) = b*b^(n-3) [3! * (n-3)! ] * b*b^(n-3) .............3 2(n - 2) a ....8 ------------- = ---- 6*b.............3 => a/b = 8/(n -2) ------5) from eqn 4) and 5) we get 10/(n-1) = 8/(n-2) =>10(n - 2) = 8(n -1) => 10n - 20 = 8n - 8 => 2n = 12 => n = 6 now eqn 1 becomes 6C1 x a x b^5 = 12 or 6ab^5 = 12 ab^5 = 2 we know from a/b = 10/(n - 1) => a/b = 10/5 = 2 or a = 2b (2b)*b^5 = 2 b^6 = 1 => b = 1 hence a = 2*1 = 2 so the expansion is (1+2)^6 = 6C0*(1^6)*(2^0) + 6C1*(1^5)*(2^1) + 6C2*(1^4)*(2^2) + 6C3*(1^3)*(2^3) + 6C4*(1^2)*(2^4) + 6C5*(1^1)*(2^5) + 6C6*(1^0)*(2^6) or (1+2)^6 = 1 + 6*2 + 15*1*4 + 20*1*8 + 15*1*16 + 6*1*32 + 1*1*64 (1+2)^6 = 1 + 12 + 60 + 160 + 240 + 192 + 64 so 2nd term = 12, 3rd term = 60 , 4th term = 160 Hope this helped Vick
Probability and Binomial Distributions?
1)Suppose you buy four boxes of the Kraked Korn Cereal. Remember that each box has an equal probability of containing any one of the seven collector cards a)what is the probability of getting 1)four identical cards 2)three identical cards 3)two identical and two different cards 4)two pairs of identical cards 5)four different cards 2)The local newspaper publishes a trivia contest with 12 extremely difficult questions, each having 4 possible answers. You have no idea what the correct answers are,so you make a guess for each question a)explain why this situation can be modelled by a binomial distribution b)use a simulation to predict the expected number ofcorrect answers c)verify your prediction mathematically d)what is the probability that you will get atleast 6 answers correct e)what is the probability that you will get fewer than 2 answers correct f)describe how the graph of this distribution would change if the number of possible answers for each question incr. or decr
Any mathematical uses of Pascal's Triangle?
Pascal's triangle comes up many times in different parts of mathematics. The most simple and easiest use is in expanding coefficents of a binomail. More specifically, it is the Binomial Theorem which states the following equality: (a + b)^n = nC0 a^n b^0 + nC1 a^(n-1) b^1 + ... nCi a^(n-i) b^i + ... + nCn a^0 b^n where nCi is the equivalent of n choose i which from combanitorics is equal to n!/i!(n-i)!. Pascal's triangle is constructed as 0C0 1C0 1C1 2C0 2C1 2C2 and so on. Equivalently it can be written in its integer form as 1 1 1 1 2 1 and so on. The really important thing to notice about Pascal's triangle is that it displays what is called Pascal's identity: nCk = (n-1)Ck + (n-1)C(k-1) which can be used to prove some other useful combinatorics identities, such as the lower limit identity. There is a much more in depth description of Pascal's triangle's uses in more advanced mathematics in the sources link.
Normal and Poisson Distribution?
1. Normal question. When you look up normal variable tables you use the formula z = (x - m)/s. Since you do not know m (mean) and s (standard deviation) you must use this formula in reverse. Tables tell you that a probability of 0.67 comes from z = 0.44 and that a probability of 0.937 comes from z = 1.53. This allows you to make a pair of simultaneous equations for m and s which you can solve. I'll leave you to do this. You should get an m value, which is the same as E(X), of a little over 9. 2. Poisson question. You can just use Poisson tables for the first part or put m = 4 and r = 0, 1, 2, 3, 4, 5 into the formula and subtract the total for r = 0, 1, 2 from the total for r up to 5. For part (b) you need to work out 1 - P(r = 0 or 1 or 2) to give you the probability of 3 or more. Then you can see what proportion of this is your answer to part (a). [I'm assuming that you missed equality from the sign just before the 5 in part (b)].