Determine the number of distinguishable permutations:
a. AABBC
b. AAABBBCCCD
c. ABCC
d. ABCCDEEF
The correct answer and explanation is:
Let us tackle each part step by step and then provide the explanation.
a. AABBC
Total letters: 5
Repeated letters: A appears twice, B appears twice
Number of distinguishable permutations:
5!2!2!=1204=30\frac{5!}{2!2!} = \frac{120}{4} = 30
b. AAABBBCCCD
Total letters: 10
Repeated letters: A appears 3 times, B appears 3 times, C appears 3 times, D appears once
Number of distinguishable permutations:
10!3!3!3!=36288006×6×6=3628800216=16800\frac{10!}{3!3!3!} = \frac{3628800}{6 \times 6 \times 6} = \frac{3628800}{216} = 16800
c. ABCC
Total letters: 4
Repeated letters: C appears twice
Number of distinguishable permutations:
4!2!=242=12\frac{4!}{2!} = \frac{24}{2} = 12
d. ABCCDEEF
Total letters: 8
Repeated letters: C appears twice, E appears twice
Number of distinguishable permutations:
8!2!2!=403204=10080\frac{8!}{2!2!} = \frac{40320}{4} = 10080
Explanation (300 words)
In combinatorics, when calculating the number of distinguishable permutations of letters in a word, the key idea is to account for identical letters that make some arrangements indistinguishable from others. If all letters were unique, then the total number of permutations would simply be the factorial of the total number of letters. For instance, a word with five unique letters has 5! permutations which equals 120. However, when some letters repeat, this creates redundancy in the arrangements because swapping identical letters does not produce a new distinct outcome. To correct for this redundancy, divide the total factorial by the factorial of each repeated letter’s count.
For example, in the word AABBC, the total number of letters is five but the As and Bs repeat. The two As can be swapped with each other without creating a new unique arrangement. Similarly, the two Bs add another level of duplication. Therefore, the total permutations are 5! divided by 2! for the As and 2! for the Bs. This principle extends to larger examples. In AAABBBCCCD, three different letters repeat three times each. Each of these repeated blocks must be accounted for by dividing by 3! for each. This general rule works for any combination: count the total letters, compute the factorial, then divide by the factorial of the counts of each repeated letter. This process ensures that only the truly distinguishable permutations are counted. This is widely used in probability, word problems, and other areas where identical objects are involved.