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Difficulty:
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Question Stats:
76% (01:39) correct
24%
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The letters K, O, O, K, and Y can be rearranged to make 5 letter strings, such as OOKKY or KOOKY. How many distinct 5 letter Strings can be created in which the two K's are NOT adjacent?
A. 9
B. 18
C. 30
D. 36
E. 60
A. 9
B. 18
C. 30
D. 36
E. 60
21 Jun 2025, 10:17
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Divyanshu14
Total arrangements = 5!/(2!2!) = 30
Arrangements where the Ks are adjacent: (KK)(O)(O)(Y) = 4!/2! = 12
Not adjacent= 30-12 = 18
GraemeGmatPanda
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26 Jun 2025, 09:56
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First count the total number of distinct arrangements of the letters without any restrictions.
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
We then have to remove the duplicates coming from the two Ks and two O2
\(2! = 2 \times 1 = 2\)
\(2! = 2 \times 1 = 2\)
\(\frac{5!}{2! \times 2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30\)
Then we can count the number of arrangements in which the two K's are adjacent by treating the pair "KK" as a single entity.
\(4! = 4 \times 3 \times 2 \times 1 = 24\)
Again we have to remove the duplicates
\(2! = 2 \times 1 = 2\)
\(\frac{4!}{2!} = \frac{24}{2} = 12\)
Subtract the number of arrangements with adjacent K's from the total to get the number with K's not adjacent.
\(30 - 12 = 18\)
Answer 18
Hope this helps!
ʕ•ᴥ•ʔ
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
We then have to remove the duplicates coming from the two Ks and two O2
\(2! = 2 \times 1 = 2\)
\(2! = 2 \times 1 = 2\)
\(\frac{5!}{2! \times 2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30\)
Then we can count the number of arrangements in which the two K's are adjacent by treating the pair "KK" as a single entity.
\(4! = 4 \times 3 \times 2 \times 1 = 24\)
Again we have to remove the duplicates
\(2! = 2 \times 1 = 2\)
\(\frac{4!}{2!} = \frac{24}{2} = 12\)
Subtract the number of arrangements with adjacent K's from the total to get the number with K's not adjacent.
\(30 - 12 = 18\)
Answer 18
Hope this helps!
ʕ•ᴥ•ʔ
