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30 Dec 2021, 06:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
57% (01:42) correct
43%
(01:37)
wrong
based on 126
sessions
History
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Time
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Not Attempted Yet
In how many ways can the letters O, R, D, E, and R be arranged so that there are at least two letters between two "R"s?
(A) 12
(B) 18
(C) 24
(D) 36
(E) 60
(A) 12
(B) 18
(C) 24
(D) 36
(E) 60
30 Dec 2021, 06:39
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If there are at least two letters between the R's, our word can only look like RXXXR, RXXRX, or XRXXR. In each of these 3 cases, we'll have 3! = 6 ways to fill in the remaining letters, so the answer is 18.
If you wanted to confirm that answer, or practice a wider variety of problem, you could solve 'backwards', by counting how many words we could make with no restrictions, then removing the words we do not want to count, though this definitely is a less efficient approach here. Since we have 2 R's, we could make 5!/2! = 60 words in total using these 5 letters, if we had no restrictions. Now if we want to divide those 60 possibilities into cases, where the R's are separated by differing numbers of letters:
• if the R's should be separated by 0 letters, so if "RR" needs to appear together in a word, then we'd really be arranging the four "letters" O, D, E, and RR, and we could make 4! = 24 words
• if the R's should be separated by 1 letter, then the first R can be the first, second or third letter in the word (i.e. our word can look like RXRXX, XRXRX or XXRXR), and in each case we'll have 3! = 6 ways to fill in the remaining letters, so in 3*6 = 18 words the R's will be separated by a single letter.
Those two cases are the only cases we do not want to count to solve the original problem, so there must be 60 - 24 - 18 = 18 words we can make where the two R's are separated by at least two letters.
If you wanted to confirm that answer, or practice a wider variety of problem, you could solve 'backwards', by counting how many words we could make with no restrictions, then removing the words we do not want to count, though this definitely is a less efficient approach here. Since we have 2 R's, we could make 5!/2! = 60 words in total using these 5 letters, if we had no restrictions. Now if we want to divide those 60 possibilities into cases, where the R's are separated by differing numbers of letters:
• if the R's should be separated by 0 letters, so if "RR" needs to appear together in a word, then we'd really be arranging the four "letters" O, D, E, and RR, and we could make 4! = 24 words
• if the R's should be separated by 1 letter, then the first R can be the first, second or third letter in the word (i.e. our word can look like RXRXX, XRXRX or XXRXR), and in each case we'll have 3! = 6 ways to fill in the remaining letters, so in 3*6 = 18 words the R's will be separated by a single letter.
Those two cases are the only cases we do not want to count to solve the original problem, so there must be 60 - 24 - 18 = 18 words we can make where the two R's are separated by at least two letters.
30 Dec 2021, 12:16
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Bunuel
Case1: R_ _ _R
So for first position we have to select between 2R's = 2C1 = 2 and last digit is fixed so its 1
Second position we can select from 3 letters =3
third from 2 letters = 2
fourth one letter one position = 1
So cases = 2*3*2*1 = 12.
Case2: R_ _R_
So for first position we have to select between 2R's = 2C1 = 2 and second last digit is fixed so its 1
Second position we can select from 3 letters =3
third from 2 letters = 2
fifth one letter one position = 1
So cases = 2*3*2*1 = 12.
Similarly, there is one more case
Case3: _R_ _R
Again here 12 combinations are possible
So total there 12*3 = 36 combinations possible. D
