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

First, we list the letters in the word ‘SYLLABUS’. We have $$S,\,Y,\,L,\,L,\,A,\,B,\,U,\,S$$. So, the multiset of letters contains $$S$$ twice, $$L$$ twice, and $$Y,\,A,\,B,\,U$$ each once.

We want to form 4-letter words in which exactly two letters are alike (a pair) and the remaining two letters are different from each other and from the repeated letter. Symbolically we need a pattern of the type $$a\,a\,b\,c$$ with $$b\neq c\neq a$$.

Step 1: choose the letter that appears twice. Only letters that occur at least twice in ‘SYLLABUS’ can serve this role. Those are $$S$$ and $$L$$. Hence, the repeated letter can be chosen in $$2$$ ways.

Step 2: choose the two distinct letters $$b$$ and $$c$$ from the remaining letters, making sure they are different from each other and from the repeated letter.

If the repeated letter is $$S$$, the available distinct types are $$\{Y,\,L,\,A,\,B,\,U\}$$, i.e. 5 different letters. The number of ways to pick any two of them is the combination $$\binom{5}{2}=10.$$

If the repeated letter is $$L$$, the remaining distinct types are $$\{S,\,Y,\,A,\,B,\,U\}$$, again 5 letters, giving $$\binom{5}{2}=10$$ ways.

So, the total number of choices for the 4-letter multiset $$\{a,a,b,c\}$$ is $$2\times 10 = 20.$$

Step 3: arrange the chosen letters. For any specific selection $$a,a,b,c$$, we must count all permutations of these four symbols where the two $$a$$’s are indistinguishable but $$b$$ and $$c$$ are distinguishable. The permutation formula for a multiset says $$\text{Number of permutations}=\dfrac{4!}{2!}=12,$$ because we divide by $$2!$$ to account for the identical pair of $$a$$’s.

Step 4: multiply the number of selections by the number of arrangements: $$20 \times 12 = 240.$$

So, the answer is $$240$$.