Worked solution
Choose three gaps for the three vowels
From the 6 letters of the word SALOON, find the
(i) number of possible arrangements if no two vowels are next to each other, marks
(ii) number of possible arrangements if the two O’s are separated, marks
(iii) number of 3-letter code words that can be formed. marks
[2011/SRJC/II/6]
CJC · Review problems · Question 4
Answer
(i) 72, using corrected gap factor ; (ii) 240; (iii) 72.
(i) Three consonants, then three vowel gaps
Arrange S, L and N in ways. There are four gaps, including the two ends. Choose three different gaps and arrange A, O, O in them: . The source diagram shows these four gaps and all three vowels. Its printed factor is a typographical error: that product would be 108, not the printed 72. The corrected gap count preserves the intended answer 72.
(ii) Method 1: subtract adjacent O’s
There are unrestricted arrangements. If the two O’s form one block, the five objects are distinct, giving . Hence .
(ii) Method 2: insert O’s into different gaps
Arrange S, A, L and N in ways. Their five gaps can receive the two identical O’s in ways: . This restriction allows A to be adjacent to O, unlike part (i).
(iii) Split by repetition pattern
Three different letter types give . The only possible repeat is O, O with one of S, A, L or N: . Thus .
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Common mistakes
The source prints an incorrect gap-selection factor in(i). Separating all vowels is a stronger condition than separating just the two O’s.
Solution notes
All asks, source diagram, both(ii) methods and code-word cases retained. Source (i) visibly prints 4 C2 but its intended/result 72 needs 4 C3; this is explicitly qualified, with original native text unchanged.