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Posted: 2/23/2002 4:04:37 AM EDT
I had a math teacher tell me that from the letters and numbers, abc123; There are over 17 million combinations that can be used. She asked me how they came up with that figure. Does anyone have the answer? signed Einstein. ;)
Link Posted: 2/23/2002 4:23:35 AM EDT
first of all, your teacher's wrong (assuming she's not counting "aaaaaa" etc. as a combination. (Even then, I dont think she'd be right.) The actual number is 720 different combinations. The way to figure it out is by doing a permutation of 6 on 6. What this is, for those who don't know, is: suppose you had those 6 numbers in a hat and drew one after the other out until youve drawn all 6. In this case, the order matters, which means "cba321", "b2c3a1," "cab132" etc. are all completely different outcomes. To count this, you do a permutation of 6 out of 6 (meaning: of 6 choices in the "hat", you draw 6 at random). The math for this is 6!/(6-6)!, or 6!/0!, or simply 720. Hope it helps...I think Im right, I may be missing some important statistical principle, but as far as I know this is the way.
Link Posted: 2/23/2002 4:36:02 AM EDT
I think your teacher didn't explain it quite well enough. I think what you're after is not a combination or permutation, but something akin to the total number of license plates possible given 3 letters followed by 3 digits. If this is what she means, the answer is 26x26x26x10x10x10=17,576,000 A-Z=26 0-9=10
Link Posted: 2/23/2002 4:41:55 AM EDT
abc,acb,bca,bac,cab,cba + 123,132,231,213,312,321 combine any set together and do the same thing. abc123,abc132,abc231,etc.etc. #c*12* aw hell its been too long since i was in math class to know the exact formula. assuming the characters could only be used once in any equasion, its a probabilities game randomely selected, there are 6 probabilities for c(character)1, then ehatever C is in position 1 can't be in any other position, so there are only 5 C's to choose from in the second choice. etc,etc 6*5*4*3*2*1=720 for each C1 multiplie that thimes the nomber of possable C1's 720*6=4320 the # of possable C2's is equal to 5*4*3*2*1=120 for each C2 multiplie for each possable C2 120*5=600 keep doing this and you come up with 4*3*2*1=24*4=96 3*2*1=6*3=18 2*1=1*2=2 1=1*1=1 add 1+2+18+96+600+4320=5037 permutations. this is for ONE possable C1 so the possabilities of any set of # starting with A = 1 in 5037 multiplie this # times the # of possable C1's and you come up with 5037*6=30222 so the possabilitie of any permutation begining with A3 (for example) is 1 in 30222 multiplie this # times the # of possable C2's and you come up with 30222*5=151110 keep doing this and this is what you get. 151110*4=604440*3=1813320*2=3626640*1 or maby its *6-#of previouse C's anyway you get the idea that it is a long mathematical equasion.. that you have to be einstien to really understand..
Link Posted: 2/23/2002 4:45:03 AM EDT
Originally Posted By Berserker: first of all, your teacher's wrong (assuming she's not counting "aaaaaa" etc. as a combination. (Even then, I dont think she'd be right.) The actual number is 720 different combinations. The way to figure it out is by doing a permutation of 6 on 6. What this is, for those who don't know, is: suppose you had those 6 numbers in a hat and drew one after the other out until youve drawn all 6. In this case, the order matters, which means "cba321", "b2c3a1," "cab132" etc. are all completely different outcomes. To count this, you do a permutation of 6 out of 6 (meaning: of 6 choices in the "hat", you draw 6 at random). The math for this is 6!/(6-6)!, or 6!/0!, or simply 720. Hope it helps...I think Im right, I may be missing some important statistical principle, but as far as I know this is the way.
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you are, that is the nomber of possable outcomes with C1 = A for example and all others random.
Link Posted: 2/23/2002 4:50:53 AM EDT
Originally Posted By mattja: I think your teacher didn't explain it quite well enough. I think what you're after is not a combination or permutation, but something akin to the total number of license plates possible given 3 letters followed by 3 digits. If this is what she means, the answer is 26x26x26x10x10x10=17,576,000 A-Z=26 0-9=10
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or if you cannot use the same character twice in the same sequence it would be 26*25*24*10*9*8=11232000 but thats given the full spectrum of possabilities.
Link Posted: 2/23/2002 5:05:53 AM EDT
Originally Posted By the_survivalist:
Originally Posted By mattja: I think your teacher didn't explain it quite well enough. I think what you're after is not a combination or permutation, but something akin to the total number of license plates possible given 3 letters followed by 3 digits. If this is what she means, the answer is 26x26x26x10x10x10=17,576,000 A-Z=26 0-9=10
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or if you cannot use the same character twice in the same sequence it would be 26*25*24*10*9*8=11232000 but thats given the full spectrum of possabilities.
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Yep, but that requirement wasn't specified in the original problem. :) Speaking of licence plates, I have a plate from my dad's '53 Chevy that reads "ILV469". I don't think that would make it past the thought police these days.
Link Posted: 2/23/2002 5:16:30 AM EDT
i am begining to have nightmares of my school (indoctrination camp) days, i will stop looking in this thread now.
Link Posted: 2/23/2002 6:03:43 AM EDT
Now I have a headache. Thanks. Jon "Low Wattage" Sherman
Link Posted: 2/23/2002 6:40:50 AM EDT
Mattja is right if the order is 3 letters followed by 3 numbers. If these can be in any order...then the number goes way up from there. License plates are figured this way because the order of the letters and numbers is constant.
Link Posted: 2/23/2002 7:10:36 AM EDT
Link Posted: 2/23/2002 7:20:21 AM EDT
Originally Posted By The_Beer_Slayer: If the letter A,B,C are travelling west bound on a train at 50 mph. And the numbers 1,2,3 are travelling east bound at 72 mph. How long will it take janet reno to burn your house down?
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30 seconds?
Link Posted: 2/23/2002 8:57:22 AM EDT
Obviously you've missed the whole point of this math lesson: How did you FEEL about the question? Knowing that there is no "right" or "wrong", only different shades of gray, what is YOUR answer? Now paint a picture of your thoughts of what your teacher told you. Did this math lesson trigger any latent homosexual thoughts that our team of school counselors can help you uncover? THESE are the real lessons to be learned from your math problem.
Link Posted: 2/23/2002 9:38:21 AM EDT
Originally Posted By The_Beer_Slayer: If the letter A,B,C are travelling west bound on a train at 50 mph. And the numbers 1,2,3 are travelling east bound at 72 mph. How long will it take janet reno to burn your house down?
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BWAHAHAHAHA!!!
Link Posted: 2/23/2002 10:19:11 AM EDT
Originally Posted By The_Macallan: Obviously you've missed the whole point of this math lesson: How did you FEEL about the question? Knowing that there is no "right" or "wrong", only different shades of gray, what is YOUR answer? Now paint a picture of your thoughts of what your teacher told you. Did this math lesson trigger any latent homosexual thoughts that our team of school counselors can help you uncover? THESE are the real lessons to be learned from your math problem.
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Oh man!!! Don't start with that shit. This philosophy of teaching that there are no wrong answers started my senior year of high school. In a MATH class of all places. The funny part was all of us students stood up and told the teacher that she was full of shit. In math there is only CORRECT or INCORRECT answers, no gray area. She couldn't handle that. Day after day we tried to get it through her thick fat liberal skull. Thankfully she only last half a year before we wore he out and she quit. To bad the replacement, was just as bad. So we decided to NOT to go to class, headed to the library and studied on our own. Only to show up for exams. Man we got in a boat load of trouble for that one. We all ended up in the principles office (thirty kids mind you) explaining that situation. The principle just shook his head, and agreed that we were right and allowed us to continue the rest of the year of self study. We were the only ones to get A's and B's!!
Link Posted: 2/23/2002 5:23:19 PM EDT
WTH was all that crap? There are only a couple kinds of math out there that are worth a damn.... here are two of them! [:D] Beer math is: 2 beers times 37 men = 49 cases. Body count math is: 2 guerillas plus one portable, plus 2 pigs = 37 enemy killed in action. .
Link Posted: 2/26/2002 9:49:42 AM EDT
Originally Posted By Guzzler: Oh man!!! Don't start with that shit. This philosophy of teaching that there are no wrong answers started my senior year of high school. In a MATH class of all places.
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What's that quote? Something like "You can argue religion and philosphpy all you want, but if you're building rifles or space ships then 2 plus 2 had better equal 4"
Link Posted: 2/26/2002 10:12:40 AM EDT
Originally Posted By Hoplophile: What's that quote? Something like "You can argue religion and philosphpy all you want, but if you're building rifles or space ships then 2 plus 2 had better equal 4"
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2 + 2 = 5, for sufficiently large values of 2 and sufficiently small values of 5.
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