The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word ’SYLLABUS’ such that two letters are distinct and two letters are alike, is :

Permutations and Combinations PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q20 with Solution

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Correct answer: 240

Count the letters in SYLLABUS

The word SYLLABUS has 8 letters:
$Y, L, L, A, B, U, S$$ Repeated letters are: - $S$ appears $2$ times - $L$ appears $2$ times Distinct single-occurring letters are: $$Y, A, B, U$$$
Required pattern

We need 4-letter words such that:
- two letters are alike
- the other two letters are distinct from that repeated letter and from each other
So the form is: $a,a,b,c$ where $a$ is a letter available at least twice.
Choose the repeated letter

Only $S$ and $L$ can be repeated.

Hence number of choices for the repeated letter: $2$
Choose the other two distinct letters

After fixing the repeated letter, choose $2$ distinct letters from the remaining available distinct letters.
- If repeated letter is $S$ , the remaining distinct letters are $L, Y, A, B, U$ → $5$ choices
- If repeated letter is $L$ , the remaining distinct letters are $S, Y, A, B, U$ → $5$ choices
In either case, number of ways to choose the two distinct letters is: $\binom{5}{2} = 10$
Arrange the 4 selected letters

For any chosen multiset $a,a,b,c$ , number of distinct arrangements is: $\frac{4!}{2!} = 12$
Total number of words

Therefore, $2 \times \binom{5}{2} \times \frac{4!}{2!} = 2 \times 10 \times 12 = 240$
Compare with stored answer

Our derived answer is: $240$

This matches the stored correct answer.

View written solutionFree

Correct answer: 240

Count the letters in SYLLABUS

The word SYLLABUS has 8 letters:
$Y, L, L, A, B, U, S$$ Repeated letters are: - $S$ appears $2$ times - $L$ appears $2$ times Distinct single-occurring letters are: $$Y, A, B, U$$$
Required pattern

We need 4-letter words such that:
- two letters are alike
- the other two letters are distinct from that repeated letter and from each other
So the form is: $a,a,b,c$ where $a$ is a letter available at least twice.
Choose the repeated letter

Only $S$ and $L$ can be repeated.

Hence number of choices for the repeated letter: $2$
Choose the other two distinct letters

After fixing the repeated letter, choose $2$ distinct letters from the remaining available distinct letters.
- If repeated letter is $S$ , the remaining distinct letters are $L, Y, A, B, U$ → $5$ choices
- If repeated letter is $L$ , the remaining distinct letters are $S, Y, A, B, U$ → $5$ choices
In either case, number of ways to choose the two distinct letters is: $\binom{5}{2} = 10$
Arrange the 4 selected letters

For any chosen multiset $a,a,b,c$ , number of distinct arrangements is: $\frac{4!}{2!} = 12$
Total number of words

Therefore, $2 \times \binom{5}{2} \times \frac{4!}{2!} = 2 \times 10 \times 12 = 240$
Compare with stored answer

Our derived answer is: $240$

This matches the stored correct answer.