Algebra and Trigonometry (MindTap Course List) ISBN:9781305071742 Author:James Stewart, Lothar Redlin, Saleem Watson Publisher:Cengage Learning You used combinations rather than permutations; so you are finding the number of sets (order does not matter) of 3 letters including at least one vowel, no repetitions allowed. One of your sets is {A,B,C}. However, the set {A,B,C} is shared by different codes. Codes ABC, ACB, BAC, BCA, CAB, and CBA are considered different codes, because they are different permutations of the elements of the set, and for a code order matters.Each of your sets of 3 letters generates codes,and . The result requires that it be specified that repetitions are not allowed. (If repetitions were allowed, as usually happens for codes< the result would be a higher number, but the problem would not be so interesting).That result can be reached most directly by subtracting the number of 3-consonant no-repeat codes, , from the number of 3-letter no-repeat codes, .
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How many 3-letter code words can be formed if at least one of the lett [#permalink] 03 Jan 2021, 12:50
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Question Stats: 38% (02:13) correct 62% (02:19) wrong based on 66 sessionsHide Show timer StatisticsHow many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?A. 3,380B. 6,615C. 8,315D. 10,140 E. 12,215 _________________
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How many 3-letter code words can be formed if at least one of the lett [#permalink] 03 Jan 2021, 13:28
Bunuel wrote: How many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?A. 3,380B. 6,615C. 8,315D. 10,140 E. 12,215 The problem didn't tell us that we can't repeat. And we have to confirm at least one vowel, so we have more than one vowel too. For the first position, we can take one from the 26 alphabetFor the second position, we can take from the 26 alphabets againFor the third position, we will confirm the vowel from the 5, because if we don't get any vowel in the first try. We don't need to maintain any order. Thus the ways: \(26*26*5=3,380\)The answer is A _________________
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Re: How many 3-letter code words can be formed if at least one of the lett [#permalink] 10 Jan 2021, 07:46
Bunuel wrote: How many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?A. 3,380B. 6,615C. 8,315D. 10,140 E. 12,215 I answered A but the OA is C. Would you please help me to learn. I will be obliged. _________________
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Re: How many 3-letter code words can be formed if at least one of the lett [#permalink] 28 Jun 2021, 07:03
MHIKER wrote: Bunuel wrote: How many 3-letter code words can be formed from the letters of the English alphabet, if at least one of the letters is to be chosen from the vowels a, e, i, o, and u?A. 3,380B. 6,615C. 8,315D. 10,140 E. 12,215 I answered A but the OA is C. Would you please help me to learn. I will be obliged. Total ways without restriction= 26*26*26Total ways with only consonants= 21*21*21Total ways with atleast 1 Vowel= 26^3-21^3
Re: How many 3-letter code words can be formed if at least one of the lett [#permalink] 28 Jun 2021, 07:03 |