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Using all the letters of the word ARRANGEMENT how many different words using all letters at a time can be made such that both A, both E, both R both N occur together .
asked Nov 13, 2012 at 15:33
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"ARRANGEMENT" is an eleven-letter word.
If there were no repeating letters, the answer would simply be $11!=39916800$.
However, since there are repeating letters, we have to divide to remove the duplicates accordingly. There are 2 As, 2 Rs, 2 Ns, 2 Es
Therefore, there are $\frac{11!}{2!\cdot2!\cdot2!\cdot2!}=2494800$ ways of arranging it.
answered Nov 13, 2012 at 15:43
JTJMJTJM
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The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares.
There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$
answered Nov 13, 2012 at 15:41
André NicolasAndré Nicolas
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Distinguishable Ways to Arrange the Word COMPUTER
The below step by step work generated by the word permutations calculator shows how to find how many different ways can the letters of the word COMPUTER be arranged.
Objective:
Find how many distinguishable ways are there to order the letters in the word COMPUTER.Step by step workout:
step
1 Address the formula, input parameters and values to find how many ways are there to order the letters COMPUTER.
Formula:
nPr =n!/[n1! n2! . . . nr!]Input parameters and values:
Total number of letters in COMPUTER:
n = 8
Distinct subsets:
Subsets : C = 1; O = 1; M = 1; P = 1; U = 1; T = 1; E = 1; R = 1;
Subsets' count:
n1[C] = 1, n2[O] = 1, n3[M] = 1, n4[P] = 1,
n5[U] = 1, n6[T] = 1, n7[E] = 1, n8[R] = 1
step 2 Apply the values extracted from the word COMPUTER in the [nPr] permutations equation
nPr = 8!/[1! 1! 1! 1! 1! 1! 1! 1! ]
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8/{[1] [1] [1] [1] [1] [1] [1] [1]}
= 40320/1
= 40320
nPr of word COMPUTER = 40320
Hence,
The letters of the word
COMPUTER can be arranged in 40320 distinct ways.
Apart from the word COMPUTER, you may try different words with various lengths with or without repetition of letters to observe how it affects the nPr word permutation calculation to find how many ways the letters in the given word can be arranged.
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Related Question
Find the number of different permutations of the letters in the word COMPUTER. How many words of five distinct letters can be formed?
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Video Transcript
Alright, we're asked, how many permutations are there of the letters of the word? Uh computer. First thing to note we have 12345678 letters. The second thing is to make sure that they're all distinct letters. They are all different letters. If I had a double R. Or something like that, then we have to divide out the number of repeats. But in this one they gave us kind of nice because they're all individual letters so we can just figure out our permutations. Alright, so we have eight letters and we're figuring out permutations of well all eight of them. Alright. Um For this one really it's the same thing as a factorial because how many ways could I fill in that first blank eight? I have eight letters that I could start with. How many for the second. Well, once I put one of them in the first spot, I only have seven left and then six for the next spot. And then five for the one after that. Alright. And then 4, 3 2. 1 for each of those eight spots. All right. So same thing as eight factorial. Same thing as eight P. Eight. Uh And then using your calculator. 40,320 40,320 different permutations of that word. Um If we only want to pick five of those letters out there, maybe the first five, maybe the last five maybe some combination. And still talking about permutations. And we still have eight to pick from. But now we're only going to pick five. Alright, so that is eight P. Five and for that one instead uh We are going to end up with 6,726,000 720. Alright. So significantly fewer of them. Um you know? And basically what we have is we're picking out one 234, the first five of them. So we're taking out those multiples there. So if you take that 40,000 and divide by the ones that we aren't picking from, we should end up with that same 6720. Alright.