Final answer:
The word HULLABALOO has 2 L's, 3 O's, and 2 U's, allowing a calculation of combinations using permutations of a multiset formula. The arrangements beginning with U and ending with L reduce the problem to 8 letters with repeats. Treating HU as one entity when they are together, to calculate combinations, again modifies the count of entities to arrange.
Step-by-step explanation:
The word HULLABALOO contains 10 letters with some repeating letters. To find the distinguishable arrangements, we use the formula for permutations of a multiset: \(\frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}\), where \(n\) is the total number of items to arrange, and \(n_1, n_2, \ldots, n_k\) are the counts of each unique item.
a. For the entire word HULLABALOO, it has 2 L's, 3 O's, and 2 U's. Therefore, the total number of distinguishable ways of arranging these letters is calculated as \(\frac{10!}{2! \times 3! \times 2!}\).
b. To find the number of ways that begin with U and end with L, we fix these letters at the start and end, leaving 8 spaces to fill. We will then rearrange the remaining 8 letters (including the second U and L), which will be calculated as \(\frac{8!}{3! \times 2!}\).
c. When the two letters HU are together, we treat them as a single entity. Now we have 9 entities to arrange: HU, L, L, A, B, A, O, O, O. The calculation becomes \(\frac{9!}{2! \times 3!}\).