i have this binomial theorem question -?
926159
4^n = (3+1)^n\n\nexpand the same using binomial theorem r+1st  term\n\n= 3^r(nCr)\n\nyou get the LHS immediately\n\nIf you want to derive 1st principle\nNow consider the product of these four equal factors:\n\n(x + b)(x + b)(x + b)(x + b) where x = 3 and b= 1\n\nThat is, let us expand (x + b)4.\n\nAgain, if we were to actually multiply out, then before collecting like terms, we would find 24  or 16  terms; and again every term would consist of four factors (one from each).  There will be terms such as xxxx,  xxbx,  bbxx,  and so on.  That is, there will be term in x4, x³b, x²b², xb³, and b4.  What the binomial theorem does is to tell how many terms there are of each kind.  Those are the coefficients.\n\nNow, how is a term x4 produced?   By taking x from each factor.  xxxx = x4.  There is only one such term.  The coefficient of x4 is 1.\n\nHow will a term x³b be produced?\n\n(x + b)(x + b)(x + b)(x + b)\n\nBy taking b from any one factor, and x from the remaining three:\n\nbxxx + xbxx + xxbx + xxxb\n\nTherefore the number of terms equal to x³b will be the number of ways of taking 1 thing -- letter b -- from 4.  There will be 4C1 = 4 terms equal to x³b.  On adding those like terms, we will have 4x³b.\n\nA term x²b² will come from taking b from any two factors in every possible way, and x from the remaining two.\n\nbbxx + bxbx + bxxb + xbbx + xbxb + xxbb\n\nTherefore, the number of terms equal to x²b² will be the number of ways of taking 2 things -- letter b -- from 4.   We will have 4C2x²b², that is, 6x²b².\n\nA term xb³ will come from taking b from any three factors, and x from the remaining factor; therefore, there will be 4C3 terms equal to xb³; we will have 4xb³.\n\nAnd finally, a term b4 comes from taking b from each of the four factors; there is only 1 way to do this; we will have 4C4b4 = b4.\n\nThis explains the binomial coefficients for the expansion of (x + b)n when n = 4.  They are 1  4  6  4  1.\n\nAgain, the binomial coefficients are how many terms there are of each kind.  They are none other than the combinatorial numbers nCk , where k successively takes on the value 0, 1, 2, 3, 4.\n\nThis result is general.  The binomial theorem states that in the expansion of (a + b)n, the coefficients are the combinatorial numbers nCk , where k successively takes on the value 0, 1, 2, . . . , n.  Each term in the\n  expansion will have this form:          n!      \n(n − k)! k!   an − kbk  .9
812449
1/n! - 1/(n+1)! =  (n+1)!/[(n!)(n+1)!] - n!/[(n!)(n+1)!]\n                        = 1/[(n!)(n+1)!] X {(n!)(n+1-1)}\n                        = n*n!/[(n!)(n+1)!]                                 (divide by n)\n                        = n!/[(n-1)!(n+1)!]                  (divide by n!)\n                        =  1/[(n-1)!(n+1)]
789066
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
937552
You can do this two ways...\n\n1. Find P(x=1) + P(x=2) + P(x=3).\n\n2. An easier way: find P(x >= 1) by taking P(x = 0) and subtracting that from 1.\n\nEither way, from the table, you should get the same thing.\n\nedit: That rule of thumb will work for the binomial distribution because it is discrete.  It would not for, say, continuous distributions.  A better rule of thumb might be this...\n\nP(X >= x) = 1 - P(X < x).\n\nThis will always work.  Hopefully you understand the reason behind it; the probability of an event is equal to one minus the probability of the opposite of that event.
833994
You might have seen this formula\n  n!/p!(n-p)! . What it means is : total permutations divided\nby permutations per combination.  The total number of\npermutations of 4 numbers  within 10,000 numbers(0000-9999)\nis  10000!/(10000-4)!=\n(10000)(9999)(9998)(9997).\nThe number of permutations per combination of 4 is 4!.\nSo the answer is \n(10000)(9999)(9998)(9997)/\n{(4)(3)(2)}.\nThat's (1250)(3333)(9998)(9997)= \n416,416,712,497,500
1015891
Isn't it sinx divided by 6?\nSimple just cancel 'n' from the numerator and the denominator.\nU will be left with six=6
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
1090404
look at what is added (in parentheses)\n\n2 (+1) 3 (+2) 5 (+0) 5 (+3) 8 (-1) 7 (+4) 11\n\nStarting with the first and taking every other number you have +1 +0 -1.  The other values are +2 +3 +4.  So you have two patterns interwoven.  Following those patterns will yield:\n\n2, 3, 5, 5, 8, 7, 11, 9, 14, 11, 17, 13, 20
885905
Yes, and this is because of Binomial theorem.\n\nAccording to Binomial theorem\n(a+b)^n = sum(k=0 to k=n)[(a^k) * (b^(n-k))]\n\nSo, with each term in your expantion, we observe 3 things\n\n1. Power of one of the variable increases from zero to n.\n2. Power of remaining of the variable decreases from n to zero.\n3. The coefficient of each term from starting to last will have coefficient (n perm k), where n is the power of (a+b) on RHS and k is the number of the term you are evaluating.\n\nI hope this answers your question.\n\nAll the best.
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
367272
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
282640
If you think about how they are used to find combinations and permutations, then it makes sense.  nCr = n!/r!(n-r!), or how many combinations are there of "n" things taken "r" at a time. Or nPr = n!/(n-r)! finds all the different combinations, plus the different ways each combination could be arranged.\n\nIf for instance, you had 6 items and you wanted to know how many combinations of 6 you could have, obviously the answer is only 1.  If you wanted to know how many ways you could arrange those six items you know that you'd have 6 choices for the first, 5 for the second, etc or 6*5*4*3*2*1=6!\n\nBut put it in the formula\n\nnCr would be 6C6 = 6!/6!(0!) you know that 6!/6! would be one, and you know that the answer is one, so for 6!/6!(0!) to stay one, 0! must equal one.\n\nFor permutations, the same rule applies 6P6 = 6!/(6-6!) = 6!/0! and I already showed that the answer was 6!, so again, 0! must =1.
624175
When you divide by (m-n) since m=n you are dividing by zero, which is not right!
982580
Use binomial coefficients to solve this.\n\n| n |           n!\n|    |   =  ----------\n| k |        (n-k)!k!\n\nn is the number of possibilities and k is the number of unordered outcomes.\n\nCombination of six numbers from ten possibilities.\n\n| 10 |       10!          3628800         \n|      | = ------------ = --------------- = 210\n|   6 |    (10-6)!6!     (24)(720)\n\nCombination of four numbers from ten possibilities.\n\n| 10 |\n|      |\n|   4 |\n\nYou should be able to figure out the second one on your own.\nHere's a link on factorials, just in case you don't know what they are.\n\nhttp://mathworld.wolfram.com/Factorial.html
528695
x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
584200
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
22440
Well, the power is of course 99, but if you are looking for the coefficient, it is -5050.  Since terms that are constant multiples of X^99 must be multiples of all X's except for one, one simply adds the second terms in each binomial {-1,-2,...,-100}.  The sum is -5050