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 07-30-2007 #1 RulerZK Guest   Posts: n/a Some probability doubts(questions). I want techniques/strategies/concepts, not just answers.? 1)given word "EQUATION". Taking all alphabets, no repeatation allowed. Find total no. of arrangements such that no 2 consonants r together.2)5boys, 5girls r there.Find no. of arrangements such thata)no 2girls can sit togetherb)no 2girls and also no 2boys are together.2b) rephrase: neither 2 girls nor 2 boys are togeather.
 07-30-2007 #2 Amy Guest   Posts: n/a no of arrangement if no consonants are together = E Q U A T I O NQ E T A N I O U3*5*2*4*1*3*2*1 =720 ORSince consonant= 3vowel=5No of ways = 3!*5! 720no 2 i can't figure it out
 07-30-2007 #3 Benjamin H. Guest   Posts: n/a 1) equation has 5 vowels and 3 consonants V = vowel 1V2V3V4V5V6as you can see, there are 6 places we could put the consonants such that none of them touch each other. There are 3 different consonants, we want to figure out how many different ways we can place them in those 3 slots (worded differently, take 3 out of those 6 slots), and order does matter, so we want to use the probability function nPr. n = 6, r = 3, so 6P3. To take into account the different ways in which the vowels can be arranged (your teacher was nice and gave you a word with no repeats), we multiply 6P3 by 5! (factorial). Each of the 5 vowels could be in the first spot, there are 4 left over, each of which could fill the next spot, etc. 5*4*3*2*1. I don't have a calculator on hand so you'll have to plug it in yourself. 6P3 * 5!2) a)This problem is similar to the first, but doesn't require a calculator. Like before,B = boy 1B2B3B4B5B6So we have 6 places to put 5 girls, or more simply, 6 places to not put a girl. Answer = 6b) if you didn't put a girl in spots 2-5, then two boys would be next to each other. This means we can only NOT put a girl in spots 1 and 6. Answer = 2hope I didn't mess up enjoyedit: amy, the consonants can be in more than one position.edit again: oh, yea, listen to the guy below me

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