Combinations and Selections from Mixed Letter Sets: An In-Depth Analysis

In this detailed exploration, we will delve into the method of determining the number of ways to select 5 letters from a given set consisting of 5 As, 4 Bs, 3 Cs, 2 Ds, and 1 E, while adhering to specific constraints on the number of each letter. This analysis involves a thorough breakdown of the possible selections based on the combination of letters.

Understanding the Problem

The goal is to find the number of distinct ways to choose 5 letters from a mixed set of letters, where the set is defined as 5 As, 4 Bs, 3 Cs, 2 Ds, and 1 E. This problem can be approached by analyzing different cases, based on the number of each type of letter selected.

Methodology

To solve this problem, we will use a systematic approach, considering each case based on the number of a specific letter (E) selected. This methodology will help us break down the problem into more manageable sub-cases.

Case 1: No E

When no E is selected, we need to find the ways to select 5 letters from the remaining set of letters (5 As, 4 Bs, 3 Cs, 2 Ds). Let the variables be as follows:

(x_A) - number of As chosen (0 to 5) (x_B) - number of Bs chosen (0 to 4) (x_C) - number of Cs chosen (0 to 3) (x_D) - number of Ds chosen (0 to 2)

Using the equation (x_A x_B x_C x_D 5), we will enumerate all possible distributions for each value of (x_D).

Sub-case 1: (x_D 0)

(x_A x_B x_C 5) When (x_A 5), (x_B 0, x_C 0) - 1 way When (x_A 4), (x_B 1, x_C 0) - 2 ways When (x_A 3), (x_B 2, x_C 0) - 3 ways When (x_A 2), (x_B 3, x_C 0) - 3 ways When (x_A 1), (x_B 4, x_C 0) - 3 ways When (x_A 0), (x_B 5, x_C 0) - 2 ways

Total for (x_D 0): 1 2 3 3 3 2 14 ways

Sub-case 2: (x_D 1)

(x_A x_B x_C 4) When (x_A 4), (x_B 0, x_C 0) - 1 way When (x_A 3), (x_B 1, x_C 0) - 2 ways When (x_A 2), (x_B 2, x_C 0) - 3 ways When (x_A 1), (x_B 3, x_C 0) - 3 ways When (x_A 0), (x_B 4, x_C 0) - 2 ways

Total for (x_D 1): 1 2 3 3 2 11 ways

Sub-case 3: (x_D 2)

(x_A x_B x_C 3) When (x_A 3), (x_B 0, x_C 0) - 1 way When (x_A 2), (x_B 1, x_C 0) - 2 ways When (x_A 1), (x_B 2, x_C 0) - 3 ways When (x_A 0), (x_B 3, x_C 0) - 2 ways

Total for (x_D 2): 1 2 3 2 8 ways

Total number of ways for (x_E 0): 14 11 8 33 ways

Case 2: One E

When one E is selected, the equation changes to (x_A x_B x_C x_D 4). The constraints remain the same.

Sub-case 1: (x_D 0)

(x_A x_B x_C 4) When (x_A 4), (x_B 0, x_C 0) - 1 way When (x_A 3), (x_B 1, x_C 0) - 2 ways When (x_A 2), (x_B 2, x_C 0) - 3 ways When (x_A 1), (x_B 3, x_C 0) - 3 ways When (x_A 0), (x_B 4, x_C 0) - 2 ways

Total for (x_D 0): 1 2 3 3 2 11 ways

Sub-case 2: (x_D 1)

(x_A x_B x_C 3) When (x_A 3), (x_B 0, x_C 0) - 1 way When (x_A 2), (x_B 1, x_C 0) - 2 ways When (x_A 1), (x_B 2, x_C 0) - 3 ways When (x_A 0), (x_B 3, x_C 0) - 2 ways

Total for (x_D 1): 1 2 3 2 8 ways

Sub-case 3: (x_D 2)

(x_A x_B x_C 2) When (x_A 2), (x_B 0, x_C 0) - 1 way When (x_A 1), (x_B 1, x_C 0) - 2 ways When (x_A 0), (x_B 2, x_C 0) - 3 ways

Total for (x_D 2): 1 2 3 6 ways

Total number of ways for (x_E 1): 11 8 6 25 ways

Final Calculation

To find the total number of ways to select 5 letters, we add the totals from both cases:

Total for (x_E 0): 33 ways Total for (x_E 1): 25 ways

Adding these, we get: 33 25 58 ways.

Conclusion

Thus, the total number of ways to select 5 letters from the given set is 58. This comprehensive analysis covers all possible selections, ensuring a thorough understanding of the combinatorial problem at hand.