How many distinguishable permutations are there for the letters of the word INCONVENIENCING taken all together ?
15 letters in all, so 15!/something
The something is the product of all the factorials for the duplicated letters.
senselessness
Well, let's see if we can count all those distinguishable permutations without inconveniencing ourselves too much. We start with 15 letters in total: I, N, C, O, N, V, E, N, I, E, N, C, I, N, G.
Now, the word "INCONVENIENCING" has some repeated letters, so we need to take that into account. Let me put on my counting cap here...
First, let's take care of those repeated "N" and "I" letters. There are three "N"s and three "I"s. So, we divide the total number of permutations by 3! (since 3! is the number of ways we can arrange the "N"s) and then divide it again by another 3! (since 3! is the number of ways we can arrange the "I"s).
Doing some quick math, the number of permutations works out to be:
15! / (3! * 3!) = 75675600
Oh, but we're not done yet! We also need to consider the repeated vowels: "O" and "E". There are three "O"s and two "E"s. So, we divide the total number of permutations by 3! (for the "O"s) and then divide it by 2! (for the "E"s).
Taking everything into account, the final answer is:
15! / (3! * 3! * 3! * 2!) = 12612600
So, there are 12,612,600 distinguishable permutations for the letters of "INCONVENIENCING". Just remember, if you try to pronounce all those permutations, you might end up unintentionally inconveniencing yourself along the way!
To find the number of distinguishable permutations for the word "INCONVENIENCING," we first need to calculate the total number of arrangements if all the letters were unique.
The word "INCONVENIENCING" has a total of 15 letters, consisting of:
I - 3 times
N - 3 times
C - 3 times
O - 2 times
V - 1 time
E - 1 time
G - 1 time
Using the formula for the number of permutations of a set with repetition, the total number of arrangements if all the letters were unique would be:
15! / (3! * 3! * 3! * 2! * 1! * 1! * 1!) = 5,651,019,040
However, since some letters are repeated, we need to divide the total by the number of times each letter is repeated to account for overcounting. So, the number of distinguishable permutations for the word "INCONVENIENCING" is:
5,651,019,040 / (3! * 3! * 3! * 2! * 1! * 1! * 1!) = 630,116
To find the number of distinguishable permutations for the letters in the word "INCONVENIENCING," we first need to determine how many total letters are in the word.
The given word has a total of 15 letters.
We can also observe that there are repeated letters in the word: N (appears 3 times), I (appears 2 times), C (appears 2 times), and E (appears 2 times).
To find the number of distinguishable permutations, we can use the formula for permutations with repetitions, which is:
n! / (n1! * n2! * ... * nk!)
where:
n = total number of objects (letters in this case)
n1, n2,... nk = number of repetitions for each object
So, in this case, we have:
n = 15
n1 = 3 (for the letter N)
n2 = 2 (for the letter I)
n3 = 2 (for the letter C)
n4 = 2 (for the letter E)
Using the formula, we can calculate the number of distinguishable permutations:
15! / (3! * 2! * 2! * 2!) = 1,299,132,800
Therefore, there are a total of 1,299,132,800 distinguishable permutations for the letters in the word "INCONVENIENCING" taken all together.