Answers

8 answer(s)

Reply	Tim5 says: The answer is 20,872,566,000 combinations. To get the answer you need to multiply 55 by itself 6 times, however since you stated no repeating numbers you would need to subtract 1 from subsequent digits because each time there is one less number that could be used in the combination. Therefore: 55 x 54 x 53 x 52 x 51 x 50 = 20,872,566,000 Hope this helps.  (1 vote) average rating:4 Tags: numbers, math.
	posted on 2010-11-19 00:43:58 | Report abuse

Reply	petethebloke says: Is there a £15m jackpot for getting this question right?  (no votes) Tags: numbers, math.
	posted on 2010-11-19 09:29:39 | Report abuse

Reply	Jon-Richfield says: That question could be read several ways. Even without Pete's acquisitive eye for jackpots, I find it challenging. What do you mean by no repeating? Would 839404 count? How about 425683? In fact, how about 333333? (threes from 33, 23, 13, 3, 43). Or do you mean 3-13-23-33-43-53 would be a valid combination. If so, then the number is 55!/49!/6! = 28 989 675 Do please explain what it is that you do mean, or we could give you thousands of different answers, each justifialble and not one of any use.  (no votes) Tags: numbers, math.
	posted on 2010-11-19 10:07:08 | Report abuse Reply petethebloke says: >Or do you mean 3-13-23-33-43-53 would be a valid combination. If so, then the number is 55!/49!/6! = 28 989 675< Depending on whether order matters.  (no votes) Tags: numbers, math. posted on 2010-11-19 11:24:07 | Report abuse Jon-Richfield says: Yes Pete, but the question spe... petethebloke says: Point taken Jon, but the subtl... Jon-Richfield says: Fair enough. That is a pretty ...

Reply	Paul_Pedant says: I was going to post all the numbers so you could count them, but Jon and Pete have put me in a huff, so I'm not going to bother now.  (no votes) Tags: numbers, math.
	posted on 2010-11-19 19:40:56 | Report abuse