What is the sum of the 10 binomial coefficients of the form C(9, k)?
605237
The sum is 2^9\n\nThis is the number of the subset of a set having n elements.
469878
10 * 9 * 8 * 7 * 6 = 30240 combinations
529699
The first thing to do is "distribute" the 9:\n\n9*x - 9*7 -5x = 2x + 10, so rewrite it:\n9x - 63 - 5x = 2x + 10, now combine the x's on the left side:\n4x - 63 = 2x + 10 (remember it's MINUS 5x), now get rid of the 2x by subtracting it from both sides (it makes zero on the right), leaving:\n\n2x - 63 = 10 and now ADD 63 to both sides:\n2x = 73... divide by 2 on both sides:\nx = 36.5\n\nDoes that help?
670692
Good question! Easy to state but unless you are familiar with this type of problem (which I was not) it is tricky. \n\nYou really want to know? The answer is:\n\n573,147,844,013,817,084,100\n\nThis is how I started. It's easy to prove it's > 2^50\n\n100/2 = 50  So you can take 2 steps at a time anywhere from 0 to 50 times. 0 would be that you climbed the 100 steps 1 at a time. If we let r equal the number of times (from 1 to 50) that you take 2 steps at a time, and for the sake of simplicity we will assume that we can only take 2 steps either from the 0th step or from an even step, then for each value of r the number of distinct ways of doing it are given by the binomial coefficient nCr = n!/(n-r)!r!. So 50C1 = 50; 50C2= 1225; 50C3 = 19,600 etc..  \nNote: in reality there are 98 ways you can combine 1 two-step with 98 1-steps so 50C1 = 50 is too low. However,let's continue. \nWe have to sum up all of these coefficient values.\nBut this is not hard to do because the sum of nCr for r= 1 to n is 2^n (2 raised to the power of n). I am not going to prove this; that's already been done. but you get the idea from the following expansions:\n\n2^3 = (1+1)^3 = 1+3+3+1 =8\n2^4 = (1+1)^4 = 1+4+6+4+1 =16\n2^5 = (1+1)^5 = 1+5+10+10+5+1 =32 and so on...\n\nWhen we use (1+1)^n as a specific case of (a+b)^n we just get the binomial coefficients and we that they sum to a power of 2. So we know the answer is greater than 2^50.\nYou will be closer if you add 1+98C1+96C2+94C3+...+54C48+52C48+50C49+1\n\nBut there is still a problem..because you have to subtract some because of starting on an odd step..\n\nAnyways, I realized it was problem in restricted partition compositions so I went to this site and they had a calculator and I used it. That's how I got the answer lol!\n\nhttp://www.btinternet.com/~se16/js/partitions.htm
483173
((6p - 18)/(9p)) / ((3p - 9)/(p^2 + 2p))\n((6(p - 3))/3(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) * ((p(p + 2))/(3(p - 3))\n(2p(p - 3)(p + 2))/((9p)(p - 3))\n(2p(p + 2))/(9p)\n(2(p + 2))/9\n(2p + 4)/9
293335
8+9+6+(1+2+3+4+5+7)^0 = 24\n\nX^(0) = 1
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
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
294753
Write 9/[n(n-1)]=9/(n-1)-9/n. The sum is then a telescoping series, so the sum from n=2 to n=9 will be\n9/1-9/2+2/2-9/3....+9/8-9/9=9-1=8.\n\nThe nth partial sum will be 9-9/n, which has a limit\nof 9 as n-->infty. Thus, the sum of the infinite series is 9.
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
724609
1000+40+1000+30+1000+20+1000\n+10+1000+0=5100\n\nor was there something else?
689300
Q98 = 1,1,2,3,5 \n\nP.S: Looks like the fibonacci series
3866
Here's the best I can do. The answer I got is:\n\n24x^4 - 10x^3 + 690x^2 + 3190x = 1217
1182929
7x+27 over (x+9)(x-9)
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
283337
FORTY\n+ TEN\n+ TEN\n--------\nSIXTY\n\n\n29786\n+ 850\n+ 850\n--------\n31486
400429
10^9, or 1,000,000,000