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Step-by-Step Solution
Step 1: Identify the distinct letters and their counts
The word "SYLLABUS" consists of the following letters and their frequencies:
S: 2 times
L: 2 times
Y: 1 time
A: 1 time
B: 1 time
U: 1 time
Step 2: Choose which letter will appear twice
We need exactly two letters to be the same in the 4-letter arrangement. From "SYLLABUS," the letters that appear more than once are S and L. Therefore, there are 2 choices for the letter that will appear twice (either S or L). In notation, this is:
\binom{2}{1}
Step 3: Choose the other two distinct letters
After choosing the repeated letter, we must pick two additional letters, each different from each other and also different from the repeated letter. Excluding the chosen repeated letter, we have 5 distinct letters left (because from S, L, Y, A, B, U, one is used twice). We need to choose 2 out of these 5 distinct letters:
\binom{5}{2}
Step 4: Arrange the chosen letters
We now have a total of 4 letters: two identical ones and two distinct ones. The number of ways to arrange these 4 letters, where 2 of them are identical, is:
\frac{4!}{2!}
This accounts for the fact that swapping the two identical letters does not create a new arrangement.
Step 5: Calculate the final number of arrangements
Combine all choices and multiply:
\binom{2}{1} \times \binom{5}{2} \times \frac{4!}{2!}.
First, compute each part:
\binom{2}{1} = 2
\binom{5}{2} = 10
\frac{4!}{2!} = \frac{24}{2} = 12
Now multiply:
2 \times 10 \times 12 = 240.
Final Answer
Hence, the number of 4-letter arrangements with two identical letters and two distinct letters, drawn from "SYLLABUS," is \boxed{240} .
