Question

If all \(C_{s}^{\prime}\) occur together and all \(U_{3}^{\prime}\) also occur together, then how many arrangements are possible of the waty SUCCESSFUL? (a) 5745 (b) 2760 (c) 6720 (d) 5432

Short answer

Answer: None of the given options are correct. The correct answer is 59040 arrangements.

Step-by-step solution

Arrange the Objects

We are considering the word as "S\((CC)\)(UU)\(_{3}EL\)". Hence, there are 7 objects to arrange, and since frequencies of S, E, L are 1 each, we have 7! arrangements.

Distribute the \(C_{s}^{\prime}\) and \(U_{3}^{\prime}\) within the Arrangements

The \(C_{s}^{\prime}\) should be placed together. There are 2! ways to arrange the \(C_{s}^{\prime}\) within themselves. Similarly, there are 3 \(U_{3}^{\prime}\), which should be placed together. The 3 \(U_{3}^{\prime}\) can be arranged within themselves in 3! ways.

Calculate the Total Number of Arrangements

To find the total number of arrangements, we multiply the arrangements in step 1, step 2, and step 3: Total Arrangements = (7!) x (2!) x (3!) = 5040 x 2 x 6 = 60480 However, there is a case when we overcount. The overcounting occurs when we have both the \(C_{s}^{\prime}\) and the \(U_{3}^{\prime}\) together. In this case, we are considering them as a single block "\((CC)(UU)U\)" with 4 objects to arrange. We have 5! ways to arrange them. Next, we calculate the arrangements inside this overcounting case. In this block, we have 2! arrangements for the \(C_{s}^{\prime}\) and 3! arrangements for the \(U_{3}^{\prime}\), resulting in a total of (5!) x (2!) x (3!) = 120 x 2 x 6 = 1440 arrangements. Now we need to subtract this overcounting from our previous calculation: Final Arrangements = Total Arrangements - Overcounting = 60480 - 1440 = 59040 However, in the multiple-choice options, we don't have this count as an answer. So, it seems that there is an error in either the problem statement or the options provided. Therefore, none of the options (a), (b), (c), or (d) are correct.

Key concepts

Permutations

Permutations deal with the arrangements of objects where the order does matter. Think of it like organizing books on a shelf; changing the order changes the arrangement. In our exercise, the word 'SUCCESSFUL' poses a unique challenge because we're treating certain letters as if they must stick together—a common variation within permutations.

To grasp the foundation, let's consider a simpler scenario: arranging three distinct letters, A, B, and C. We could organize them in six different ways (ABC, ACB, BAC, BCA, CAB, CBA). This number of arrangements is described by the permutation formula for 'n' distinct objects, which is represented by 'n' factorial, written as 'n!'.

Factorial

The factorial, represented by the exclamation point '!', is a function that multiplies a number by all the positive integers below it. For example, '5!' equals 5 x 4 x 3 x 2 x 1, which is 120. In our exercise, we encountered factorials during the calculation of possible arrangements. Factorials play a crucial role in calculating permutations because they provide the total number of ways to arrange 'n' distinct items.

For a quick insight, let's look at the factorial of the number 3, noted as '3!'—it's simply 3 x 2 x 1, which gives us 6. That ties back to our earlier example with the three letters: there are 6 permutations for arranging three distinct objects. Factorials grow rapidly with larger numbers, so calculating them is essential to accurately determine permutations in combinatorics.

Arrangement of Objects

Arranging objects is a key focus in combinatorics, particularly in permutations. But real-world items aren't always distinct; sometimes they repeat, affecting the total arrangements. In our exercise, we approach 'SUCCESSFUL' by viewing repeating letters as single entities, a technique often used to simplify permutation problems.

When arranging objects, it’s essential to distinguish between cases where repetition is allowed and cases where it isn’t. If we treat 'CC' and 'UUU' in 'SUCCESSFUL' as individual blocks, we effectively reduce the problem to arranging fewer distinct objects. This helps avoid overcounting scenarios where repetitive objects are seen as interchangeable, a common pitfall in combinatorial exercises.