Who can get this?! This is merely just for fun. Find as much 4 number combinations as you can using the numbers 0-9, with the combinations starting with the number 5. (i.e.: 5647, 5382, 5194, 5029, and so much more!) There are FOURTY IN TOTAL. The winner will get 10 points.
200513
5012. 5013. 5014. 5016. 5017. 5018. 5019. 5021. 5023. 5024. 5026. 5027. 5028. 5029. 5031. 5032. 5034. 5036. 5037. 5038. 5039. 5041. 5042. 5043. 5044. 5046. 5047. 5048. 5049. 5061. 5062. 5063. 5064. 5067. 5068. 5069. 5071. 5072. 5073. 5074. 5076. 5078. 5079. 5081. 5082. 5083. 5084. 5086. 5087. 5089. 5091. 5092. 5093. 5094. 5096. 5097. 5098. 5102. 5103. 5104. 5106. 5107. 5108. 5109. 5123. 5124. 5126. 5127. 5128. 5129. 5130. 5132. 5134. 5136. 5137. 5138. 5139. 5140. 5142. 5143. 5146. 5147. 5148. 5149. 5160. 5162. 5163. 5164. 5167. 5168. 5169. 5170. 5172. 5173. 5174. 5176. 5178. 5179. 5180. 5182. 5183. 5184. 5186. 5187. 5189. 5190. 5192. 5193. 5194. 5196. 5197. 5198. 5201. 5203. 5204. 5206. 5207. 5208. 5209. 5213. 5214. 5216. 5217. 5218. 5219. 5230. 5231. 5234. 5236. 5237. 5238. 5239. 5240. 5241. 5243. 5246. 5247. 5248. 5249. 5260. 5261. 5263. 5264. 5267. 5268. 5269. 5270. 5271. 5273. 5274. 5276. 5278. 5279. 5281. 5283. 5284. 5286. 5287. 5289. 5290. 5291. 5293. 5294. 5296. 5297. 5298. 5301. 5302. 5304. 5306. 5307. 5308. 5309. 5310. 5312. 5314. 5316. 5317. 5318. 5319. 5320. 5321. 5322. 5324. 5325. 5326. 5327. 5328. 5329. 5340. 5341. 5342. 5346. 5347. 5348. 5349. 5360. 5361. 5362. 5364. 5367. 5368. 5369. 5370. 5371. 5372. 5374. 5375. 5376. 5378. 5379. 5380. 5381. 5382. 5384. 5386. 5387. 5389. 5390. 5391. 5392. 5394. 5396. 5397. 5398. 5401. 5402. 5406. 5407. 5408. 5409. 5410. 5412. 5413. 5416. 5417. 5418. 5419. 5420. 5421. 5423. 5426. 5427. 5428. 5429. 5430. 5431. 5432. 5436. 5437. 5438. 5439. 5460. 5461. 5462. 5463. 5467. 5468. 5469. 5470. 5471. 5472. 5473. 5476. 5478. 5479. 5480. 5481. 5482. 5483. 5486. 5487. 5489. 5490. 5491. 5492. 5493. 5495. 5496. 5497. 5498. 5601. 5602. 5603. 5604. 5607. 5608. 5609. 5610. 5612. 5613. 5614. 5617. 5618. 5619. 5620. 5621. 5623. 5624. 5627. 5628. 5629. 5630. 5631. 5632. 5634. 5637. 5638. 5639. 5640. 5641. 5642. 5643. 5647. 5648. 5649. 5670. 5671. 5672. 5673. 5674. 5678. 5679. 5680. 5681. 5682. 5683. 5684. 5685. 5687. 5689. 5690. 5691. 5692. 5693. 5694. 5697. 5698. 5701. 5702. 5703. 5704. 5706. 5708. 5709. 5710. 5712. 5713. 5714. 5716. 5718. 5719. 5720. 5721. 5723. 5724. 5726. 5728. 5729. 5730. 5731. 5732. 5734. 5736. 5738. 5739. 5740. 5741. 5742. 5743. 5746. 5748. 5749. 5760. 5761. 5762. 5763. 5764. 5768. 5769. 5780. 5781. 5782. 5783. 5784. 5786. 5789. 5790. 5791. 5792. 5793. 5794. 5796. 5798. 5801. 5802. 5803. 5804. 5806. 5807. 5809. 5810. 5812. 5813. 5814. 5816. 5817. 5819. 5820. 5821. 5823. 5824. 5826. 5827. 5829. 5830. 5831. 5832. 5834. 5836. 5837. 5839. 5840. 5841. 5842. 5843. 5846. 5847. 5849. 5860. 5861. 5862. 5863. 5864. 5865. 5867. 5869. 5870. 5871. 5872. 5873. 5874. 5876. 5879. 5890. 5891. 5892. 5893. 5894. 5896. 5897. 5901. 5902. 5903. 5904. 5906. 5907. 5908. 5910. 5912. 5913. 5914. 5916. 5917. 5918. 5920. 5921. 5923. 5924. 5926. 5927. 5928. 5930. 5931. 5932. 5934. 5936. 5937. 5938. 5940. 5941. 5942. 5943. 5946. 5947. 5948. 5960. 5961. 5962. 5963. 5964. 5967. 5968. 5970. 5971. 5972. 5973. 5974. 5976. 5978. 5980. 5981. 5982. 5983. 5984. 5986. 5987.
457503
Hold up eight fingers.  Now put down three of them.  That leaves five of them.\n\nThe answer is 5.\n\nIt is also equivalent to the number of points you lost asking this question... but I guess you are just having fun with Yahoo! Answers coming up with silly questions, aren't you.  Enjoy!
152456
The answer is Choose(n+k-1,k-1).\n\nFor example, if n=5 and k=3, the value is Choose(7,2) = 21.\n\nOne way to understand that is to take n+k-1 different pigeon holes:\n\n    O O O O O O O\n\nNow, if we choose 2 of these, say, the fourth and the sixth, like:\n    O O O X O X O\n\nthat yields the distribution as (3,1,1).  You can, I hope, see how this works in general.
137737
Could be 10,14,13. The odd terms differ by 3 (eg. 4-1,7-4), the even terms differ by 4.
486085
8\n9\n7\n4\n4\n5\n7\n7\n9\n8\n5\n4\n8\n8
668891
I tried it 3 times.  It was right all 3 times.\n\nMy numbers were:\n 11 - 2 = 9\n 99 - 18 = 81\n 54 - 9  = 45\n\nNote that all 3 are multiples of 9.  For some reason (I don't know why) doing that math gives you a multiple of 9.\n\nLets try some more\n12-3 = 9 yup\n16-7 = 9 yup\n40-4 = 36 yup\n57 - 12 = 45 yup\n\nweird....
482580
This is a combination problem, because the order of people doesn't matter. In permutations, the order does matter.\n\nThe formula for the number of combinations of k items from a set of n items total is:\n\nn! / [ k! (n - k)! ] where ! means factorial (n * (n-1) * (n-2) ... * 1)\n\nIf your case that would be 35! / (8! * 27!). Sounds like a calculator problem to me.\n\nI get 23535820 combinations.\n\nGood luck on your homework!
637928
The numbers give the next in the sequence, for example 3692 is 369*2 and that is 738 and so on.\n\n369 x 2 = 738 || 73 x 8 = 584 || 58 x 4 = 232 || 23 x 2 = 46\n\nYour final answer is 46.
104047
Each answer is 4 more than the sum of the previous two terms.  The 49 is a misprint and should be 59.\n\n(5+11) + 4 = 20\n(11+20) + 4 = 35\n(20+35) + 4 = 59\n\nso,\n\n(35+59) + 4 = 98\n(59+98) + 4 = 161
535088
The answer is 9! / (6!*3!).\n\nYou're dealing with combinations and not permutations, so we don't care which person is which... you really have nine chairs, six of which are occupied and three of which are empty. Forget about ABCDEF being the six people: Let X represent an occupied chair, and O an empty chair.\n\nIf you need to create a list of all possible combinations, just start with the X'es all to the left and the O's to the right, then do a bubble sort thingy.\n\nXXXXXXOOO\nXXXXXOXOO\nXXXXXOOXO\nXXXXXOOOX\nXXXXOXXOO\nXXXXOXOXO\nXXXXOXOOX\nXXXXOOXXO\nXXXXOOXOX\nXXXXOOOXX\netc.\n\nIf you grind your way through the whole list, you'll find there are 84 combinations. There's some neat stuff going on here with triangular numbers: you can see that there's one arrangement where the first O is in the 7th position, three where the first O is in the 6th position, six where the first O is in the 5th position... follow the pattern and you'll see that there are 1+3+6+10+15+21+28 combinations in the list. That adds up to 84.\n\nHopefully that helps!
212916
10 * 9 * 8 * 7 * 6 = 30240 combinations
224823
Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
5936
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
526617
The answer is 4.\n11 is eleven, which has 6 letters, so the next number is 6.\n6 is six, which has 3 letters, so the next number is 3.\n3 is three, which has 5 letters, so the next number is 5.\n5 is five, which has 4 letters, so the next number is 4.\n4 is four, which has 4 letters, so the next number is 4. It's all 4's from there!\n\nBy the way, 4 is the only number that has the same number of letters as its value, and if you start this sequence at any whole number, you will end up at four! Just a tidbit I knew! ^_^
64761
16,000.\n\nThree digits gives 1,000 possible combinations (from 000 to 999), times 16 counties.\n\nI note the category "college and graduate schools" -- have the colleges gone that far downhill that this is a challenge at that level??
281312
A 9 digit number has exactly 1,000,000,000 combinations if you include the code 000-00-0000 through 999-99-9999.  So we are about 1/3 of the way to running out of numbers if we are recycling.
318166
5050\n\nIf you add the numbers like this\n (1 + 100) + (2 + 99) + (3 + 98) + . . . . + (50 + 51) = ?\nthen each set adds up to 101 and there are 50 sets of numbers, so 101*50=5050