Given below are two statements :
A number of distinct 8-letter words are possible using the letters of the word SYLLABUS. If a word in chosen at random, then

Statement I: The probability that the word contains the two S's together is $$\frac{1}{4}$$

Statement II : The probability that the word begins and ends with L is $$\frac{1}{28}$$.
In the light of the above statements, choose the correct answer from the options given below

Total number of 8-letter words possible using letters S, Y, L, L, A, B, U, S is $$\dfrac{8!}{2!2!}$$

Statement I:

We can find total number of 8-letter words by considering 2 S's as a single unit and arranging 7 units, i.e. (SS), Y, L, L, A, B, U

Total number of ways = $$\dfrac{7!}{2!}$$

Required probability = $$\ \dfrac{\ \dfrac{7!}{2!}}{\dfrac{8!}{2!2!}}=\dfrac{2}{8}=\dfrac{1}{4}$$

Therefore, statement I is correct.

Statement II:

We can find total number of 8-letter words such that L's are at ends by arranging remaining 6 letters in the middle.

Total number of ways = $$\dfrac{6!}{2!}$$

Required probability = $$\ \dfrac{\ \dfrac{6!}{2!}}{\dfrac{8!}{2!2!}}=\dfrac{2}{56}=\dfrac{1}{28}$$

Therefore, statement II is correct.

The answer is option A.