What is the sum of the 10 binomial coefficients of the form C(9, k)?
397443
C(9,0)+...+C(9,9) = 2^9 = 512.
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
945958
(2n-1) * (2n-2) * (2n-3) * (2n-4)... * 3 * 2 * 1\n\nJust like any other factorial.
403326
1+2+3+4+5+6+7+8+9/3+8*2-4\n=36+3+16-4\n=51
960925
The answer you gave cannot possibly be right. Let n=1. Then 2^n * (n+2)C2 = 2^1 * 3C2 = 6, but [k=1, n]∑k²*nCk = 1² * 1C1 = 1, and 6≠1.\n\nI'll post a proof of the correct answer just as soon as I figure out what it is.\n\nEdit: after much rumination, and many used pieces of scratch paper, I have found a solution. \n\n[k=1, n]∑(k²nCk) = n(n+1)2^(n-2)\n\nIn proving this, I will assume that you are already familiar with the fact that (n+1)Ck = nCk + nC(k-1) and that [k=0, n]∑nCk = 2^n. If not, a good proof of the first is provided in the wikipedia article: http://en.wikipedia.org/wiki/Pascal%27s_rule . The second follows from the fact that this is the sum of the numbers on the nth row of my triangle, and each row has a sum twice that of the row before it (think about how each row is constructed).\n\nOur strategy in this proof will be to find a recurrence relation for each sum, and then solve this relation to obtain a general formula. First we will need to prove a lemma:\n\n[k=0, n]∑knCk = n2^(n-1)\n\nProof:\n\nFirst note that [k=0, 0]∑k0Ck = 0*0C0 = 0. Now we find a recurrence relation as follows:\n\n[k=0, n+1]∑k(n+1)Ck\nExtract the (n+1)th term:\nn+1 + [k=0, n]∑k(n+1)Ck\nExpand the binomial coefficient:\nn+1 + [k=0, n]∑k(nCk + nC(k-1))\nDistribute:\nn+1 + [k=0, n]∑knCk + [k=0, n]∑knC(k-1)\nExtract the 0th term from the right-hand sum (this term is 0):\nn+1 + [k=0, n]∑knCk + [k=1, n]∑knC(k-1)\nChange the indices of summation:\nn+1 + [k=0, n]∑knCk + [k=0, n-1]∑(k+1)nCk\nSince the term we extracted at the beginning is the nth term of the sum on the right, we can merge it into the sum:\n[k=0, n]∑knCk + [k=0, n]∑(k+1)nCk\nDistribute the k+1 term:\n[k=0, n]∑knCk + [k=0, n]∑knCk + [k=0, n]∑nCk\nSimplify:\n2[k=0, n]∑knCk + [k=0, n]∑nCk\nAnd finally:\n2[k=0, n]∑knCk + 2^n\n\nSo we are looking for an f that satisfies the recurrence relation:\n\nf(0)=0\nf(n+1) = 2f(n) + 2^n\n\nApplying this to itself:\n\nf(n+1) = 2(2f(n-1) + 2^(n-1)) + 2^n = 4f(n-1) + 2*2^n\nf(n+1) = 4(2f(n-2) + 2^(n-2)) + 2*2^n = 8f(n-2) + 3*2^n\nExtrapolating this pattern further:\nf(n+1) = 2^(n+1)f(0) + (n+1)2^n\nBut f(0)=0, so this is simply:\nf(n+1) = (n+1)2^n or f(n) = n2^(n-1)\n\nTo ensure our extrapolation is correct, we shall prove it rigorously by induction. First we note that f(0)=0, as it should. To show that n2^(n-1) satisfies f(n+1) = 2f(n) + 2^n in general, we assume it works for some n. Then:\n\nf(n+1) = 2f(n) + 2^n\nf(n+1) = 2n2^(n-1) + 2^n\nf(n+1) = n2^n + 2^n\nf(n+1) = (n+1)2^n\n\nWhich is what we get by plugging n+1 into the formula. So it works for all n.\n\nNow having proved this lemma, we proceed to [k=0, n]∑k²nCk (yes, your original formula had [k=1, n]∑k²nCk, however since the k=0 term is zero, including it does not change the sum). We'll use the function g to avoid confusing it with the previous function. Of course g(0)=0. Now we need our recurrence relation:\n\n[k=0, n+1]∑k²(n+1)Ck\nExtract the (n+1)th term:\n(n+1)² + [k=0, n]∑k²(n+1)Ck\nExpand the binomial coefficient and distribute:\n(n+1)² + [k=0, n]∑k²(nCk + nC(k-1))\n(n+1)² + [k=0, n]∑k²nCk + [k=1, n]∑k²nC(k-1)\nChange the indices of summation:\n(n+1)² + [k=0, n]∑k²nCk + [k=0, n-1]∑(k+1)²nCk\n(n+1)² is the nth term of the sum on the right, so we can merge them:\n[k=0, n]∑k²nCk + [k=0, n]∑(k+1)²nCk\nExpand the binomial:\n[k=0, n]∑k²nCk + [k=0, n]∑(k²+2k+1)nCk\nAnd distribute:\n[k=0, n]∑k²nCk + [k=0, n]∑k²nCk + 2[k=0, n]∑knCk + [k=0, n]∑nCk\nEmploying the formulas for [k=0, n]∑nCk and [k=0, n]∑knCk (the latter of which was proved as our lemma):\n2[k=0, n]∑k²nCk + 2n2^(n-1) + 2^n\nAnd simplifying further:\n2[k=0, n]∑k²nCk + (n+1)2^n\n\nSo g(n+1)=2g(n) + (n+1)2^n. Applying this to itself:\n\ng(n+1) = 2(2g(n-1) + n2^(n-1)) + (n+1)2^n = 4g(n-1) + ((n+1)+n)2^n\ng(n+1)=4(2g(n-2) + (n-1)2^(n-2)) + ((n+1)+n)2^n = 8g(n-2) + ((n+1) + n + (n-1))2^n\nExtrapolating this pattern further:
1115229
Just add them up using the calculator:\n1000 + 5999 + 9888 = 16887\n\nOr if you don't have a calculator:\n*111\n---------\n**1000\n+5999\n+9888\n=====\n16887\n=====
1182929
7x+27 over (x+9)(x-9)
638209
1, 2, 3, 6, 7, 14, 21, 42
314118
it's 8*8 + 7*7 + ... +2*2 + 1*1. \n\nso for the chessboard it would be 204. :]
250266
The fourth term in a sixth order binomial expansion has a coefficient given by 6 choose 3. This is because we need to consider all rows of Pascal's triangle beginning with row zero, column zero. 6 choose 3 falls on row 6 column 3 and is 6!/(3!*3!)=20. This factor will then be multiplied by x^3 as well as (-2y)^3. Together the fourth term will be:\n\n-160x^3y^3
758683
Hm.  There are lots of good sites on the web about Pascal's triangle and the Binomial expansion.\n\nTo understand how to get the coefficients, consider as an example trying to compute:\n\n(x + y)³\n= (x + y)(x + y)(x + y)\n= (xx + xy + yx + yy)(x + y)\n= xxx + xyx + yxx + yyx + xxy + xyy + yxy + yyy\n= xxx + (xyx + yxx + xxy) + (yyx + xyy + yxy) + yyy\n\nI deliberately didn't simplify to make it clear what is happening.\n\nNotice that all the terms in parentheses are the same (x²y and xy²), and will ultimately add together....\n\n....but also notice that you can think of these four groupings as:\n\n-- The number of ways to form a string of length 3 where each character is an x.\n-- The number of ways to form a string of length 3 where two characters are x and the other one is a y.\n-- The number of ways to form a string of length three where two characters are y and the other one is an x.\n-- The number of ways to form a string of length 3 where each character is a y.\n\nBut, when you group them to add them, order no longer makes a difference: xxy = xyx = yxx = x²y, and yyx = yxy = xyy = xy².\n\nSo to compute a particular value in the triangle in the nth row and in the mth column, what you are really trying to do is to calculate "how many strings of length n can I create using (n - m + 1) x's and (m - 1) y's"?  (the extra +1 and -1 are there so you don't have to call the first column the "0th column".)\n\nFor example, take the 4th row, 2nd column.  This would be the coefficient of x³y in the binomial expansion of (x+y)^4.\n\nThe equivalent question is:\n\n"How many strings of length 4 can you create using (4 - 2 + 1) = 3 x's and (2 -1) = 1 y?"\n\nYou could write them all out: xxxy xxyx xyxx yxxx.\n\nOr you can say, I have 4 places for the first x, 3 places for the second x, 2 places for the third x, and only one left for the y, for a total of 4! = 24 arrangements.  But since the x's can't be distinguished from each other, you have to divide by the factorial of the number of x's. 4! / 3! = 4.\n\nHopefully this is familiar to you: it's a combination.\n\nSo, in general, to compute the value in the mth column and nth row of the triangle, the formula is:\n\nC(n, m-1) = n! / [(m - 1)! (n - m + 1)!]\n\nOr if you're ok with calling the calling the first column "column zero", then it's simpler:\n\nC(n, m) = n! / [m! ( n - m)!]\n\nHope that gives you an idea.  :)
1080867
This is the Gaussian binomial coefficient [n,1] for q=2. \n\nGreetings from The On-Line Encyclopedia of Integer Sequences!
801198
k + 9 = 6k - 66\n5 k = 9+66 = 75\nk = 15
383816
!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
813994
the first one is (x+10)(x+8)\nand the second one is your second answer
795050
Well I am a little unclear with ur question!...considering the order is not important we can give itin many ways assuming we have to choose only one question(where this would be of 30 pts obviously), we can choose it in 10c1 ways (c = combination, where NcR = [N! / [(N-R)!R! and N! = N * N-1 *....*1] therefore 10c1 = 10 \n \naccordingly we choose 2 questions of these 10 as 10c2 = 45 (assuuming the weightage of questions is insiginficant but their sum =30 pts) \n\naccordingly + 10c3 + 10c4 + 10c5....10c10. which is equal to \n10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1023\n\nthere is actually a formula for this too, which is 2^n - 1 where n is the number of questions(like 10 here)
283337
FORTY\n+ TEN\n+ TEN\n--------\nSIXTY\n\n\n29786\n+ 850\n+ 850\n--------\n31486