AlbaLed Posted October 26, 2003 Share Posted October 26, 2003 These are simple ones, but usually overlooked 1. How many bit strings oflength 10 have a. exactry three 0-s ? b. same number of 0-s as 1-s? c. at least seven 1-s? d. at least three 1-s ? 2. The English alphabet contains 21 consonants and 5 vowels. How many strings of 6 lowercase letters of the English alphabet contain: a. exactly 1 vowel ? b. exactly 2 vowels ? c. at least 1 vowel ? d. at least 2 vowels? 3. What is the coefficient of x^101y^99 in the expanssion of (2x - 3y)^200? AlbaLed Quote Link to comment Share on other sites More sharing options...
jaideep Posted October 26, 2003 Share Posted October 26, 2003 1) a) 10 C 3 = 120 b) 10 C 5 = 252 c) 10 C 7 + 10 C 8 + 10 C 9 + 10 C 10 = 176 d) 2^10 - 10 C 0 - 10 C 1 - 10 C 2 = 968 2) a) 5.6.21.21.21.21.21 b) 15 . 6 C 2 . 21 . 21 . 21 . 21 c) 26^6 - 21^6 d) 26^6 - 21^6 - 5.6.21.21.21.21.21 3) 200 C 101 * 2^101 * (-3)^99 Any explaination would be highly appreciated. :) Jaideep Quote Link to comment Share on other sites More sharing options...
AlbaLed Posted October 26, 2003 Author Share Posted October 26, 2003 Originally posted by jaideep 1) a) 10 C 3 = 120 Correct !! b) 10 C 5 = 252 Correct !! c) 10 C 7 + 10 C 8 + 10 C 9 + 10 C 10 = 176 Correct !! But working with the big numbers might get you to trouble, you can recognize from above that 10 C 7 = 10 C 3, 10 C 8 = 10 C 2, 10 C 9 = 10 C 1 and 10 C 10 = 10 C 0, and start working with smaller numbers, or you can reverse the problem into bit-string of length 10 with at most three 0-s in it, which would directly yield an expression with smaller number. It's all a matter of preference. d) 2^10 - 10 C 0 - 10 C 1 - 10 C 2 = 968 2) The trick here is to note that repetitions are permitted. Let v denote number of vowels, and c number of consonants More formally for a string of length n with exatly m vowels we have v^m*(n C m)*c^(n-m) Why ????? First let's find in how many different combinations of m vowels and n-m consonants there are. This problem is dentical to the problem of finding how many bit-strings of length n contain m 0-s (or 1-s) [(n C m)]. Now for each group (consonants and vowels we have different choices 5-vowels, 21-consonants). For each group find how many permutations (allowing repetition in this case) of length m (vowels) and n-m (consontants) exist[v^m, c^(n-m)]. Does this make any sense? What I am basically doing is solving the problem piece by pience and matchin any piece with any existing problem whose solution is apparent. a) 5.6.21.21.21.21.21 Correct !! b) 15 . 6 C 2 . 21 . 21 . 21 . 21 Don't understand where 15 comes from, maybe you meant 25 ??? c) 26^6 - 21^6 Correct !!! This set of strings can be thought of as the set of all string with length 6 - the set of all string of length 6 with no vowels on them d) 26^6 - 21^6 - 5.6.21.21.21.21.21 Correct !!! This set of strings can be thought of as the set of all string with length 6 - the set of all string of length 6 with no vowels on them - the set of all string of length 6 with 1 vowel. (no vowel an 1 vowel in the string violate the request - at least two vowels) Of course you can solve this problem by counting all strings (length 6), with 2,3,4,5,6 vowels but that is quite a long way, same goes for C. 3) 200 C 101 * 2^101 * (-3)^99 Great, binomial expanssion is the right choice here Any explaination would be highly appreciated. :) Jaideep Good job Jaideep AlbaLed Quote Link to comment Share on other sites More sharing options...
jaideep Posted October 26, 2003 Share Posted October 26, 2003 Hey AlbaLed, Q2)b) Here is the explainations for the answer. You can select two vowels in 5 C 2 ways = 10 You can also select same vowel twice in 5 ways so total = 10 + 5 = 15 You can arrange these 2 vowels in a string of length 6 in 6 C 2 ways. And rest of the places can be filled by 21 choices given by consonants. Hey Alba can you explain why you think it should be 25 and not 15. Jaideep Quote Link to comment Share on other sites More sharing options...
Laks Posted October 26, 2003 Share Posted October 26, 2003 Hey Dude, You can select one vowel in 5C1 ways... agin we can choose from 5 vowels since repetitions are allowed... So we get another 5C1. Hence 5*5 i.e.25. Cheers' Laks Quote Link to comment Share on other sites More sharing options...
AlbaLed Posted October 26, 2003 Author Share Posted October 26, 2003 Think of generating all possible combinations of two vowels, with repetition, you can think of it as a 2 digit number with base 5 (number of vowels) therefore 5^2. Laks, your explanation makes sense too AlbaLed Quote Link to comment Share on other sites More sharing options...
jaideep Posted October 26, 2003 Share Posted October 26, 2003 ooops. Missed that point. Thanks for correcting me guys. :) Jaideep Quote Link to comment Share on other sites More sharing options...
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