a) In how many ways the letters of the word ALGEBRA can be arranged so that repeated letter are never together ? [2] b) Show that 12! + 23! + 34! + ... = 1 . [2] c) Prepare a Cayley's table for S =…

a) Arrangements of the word ALGEBRA with repeated letters never together

Step 1: Identify the letters and their frequencies The word ALGEBRA has 7 letters with the following repetitions:

• A: 2 times
• L: 1 time
• G: 1 time
• E: 1 time
• B: 1 time
• R: 1 time

Step 2: Total arrangements without restrictions If all letters were distinct, the total number of arrangements would be $7!$. However, since A is repeated twice, the total arrangements are: $$\frac{7!}{2!}=\frac{5040}{2}=2520$$

Step 3: Treat the repeated letters (A, A) as identical and find arrangements where they are together If the two As are treated as a single entity, we now have 6 entities to arrange: $$\text{AA, L, G, E, B, R}$$ The number of arrangements is: $$6!=720$$

Step 4: Calculate arrangements where the two A's are not together Subtract the restricted case (A's together) from the total arrangements: $$2520-720=1800$$

Final Answer: The number of ways to arrange the letters of ALGEBRA such that the repeated As are never together is 1800.

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b) Prove that $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots =1$

Step 1: Generalize the series The given series is: $$S=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots +\frac{n}{(n+1)!}$$

Step 2: Rewrite each term Notice that: $$\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{k+1}{(k+1)!}-\frac{1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$$

Step 3: Apply telescoping series logic Substitute the rewritten term into the series: $$S=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\dots$$

Step 4: Observe cancellation Most terms cancel out: $$S=\frac{1}{1!}-\underset{n\to \infty }{\lim }\frac{1}{(n+1)!}$$

Step 5: Evaluate the limit Since $\frac{1}{(n+1)!}\to 0$ as $n\to \infty$: $$S=1-0=1$$

Final Answer: $$\boxed{1}$$

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c) Cayley's table for $S=\{0,2,3\}$ under addition modulo 4

Step 1: Define the operation The set $S=\{0,2,3\}$ is closed under addition modulo 4. The operation table (Cayley's table) is constructed by computing $(a+b)\mathrm{mod}4$ for all $a,b\in S$.

Step 2: Construct the table

Explanation of entries:

• $0+0=0\mathrm{mod}4$
• $0+2=2\mathrm{mod}4$
• $0+3=3\mathrm{mod}4$
• $2+0=2\mathrm{mod}4$
• $2+2=4\equiv 0\mathrm{mod}4$
• $2+3=5\equiv 1\mathrm{mod}4$ (Note: 1 is not in $S$, but the table is still valid for the operation.)
• $3+0=3\mathrm{mod}4$
• $3+2=5\equiv 1\mathrm{mod}4$
• $3+3=6\equiv 2\mathrm{mod}4$

Correction: Since the question specifies $S=\{0,2,3\}$, but $1$ appears in the table, we must clarify that the operation is defined for all pairs in $S$, but the result may not always lie in $S$. However, if we restrict the table to results within $S$, we must exclude invalid entries. But typically, Cayley's table includes all possible results under the operation, even if they are outside the set.

Revised Table (if only results in $S$ are considered): Since $2+3=1\notin S$, the operation is not closed under addition modulo 4 for $S=\{0,2,3\}$. However, if the question expects the full table (including results outside $S$), the first table is correct.

Assuming the question expects the full table (including all results under modulo 4):

Note: If the question implies that the operation is restricted to results within $S$, then the set $S$ is not closed under addition modulo 4, and the table cannot be fully constructed within $S$. However, based on standard interpretations, the first table is provided.

Final Answer: The Cayley's table for $S=\{0,2,3\}$ under addition modulo 4 is:

| + | 0 | 2 | 3 |
|---|---|---|---|
| 0 | 0 | 2 | 3 |
| 2 | 2 | 0 | 1 |
| 3 | 3 | 1 | 2 |