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Now we address the second problem.

Permutations with Similar Elements

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labeling the letters as follows.

E 1 LE 2 ME 3 NT size 12{E rSub { size 8{1} } ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{3} } ital "NT"} {}

Since all the letters are now different, there are 7 ! size 12{7!} {} different permutations.

Let us now look at one such permutation, say

LE 1 ME 2 NE 3 T size 12{ ital "LE" rSub { size 8{1} } ital "ME" rSub { size 8{2} } ital "NE" rSub { size 8{3} } T} {}

Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are 3 ! size 12{3!} {} or 6 such arrangements. We list them below.

LE 1 ME 2 NE 3 T size 12{ ital "LE" rSub { size 8{1} } ital "ME" rSub { size 8{2} } ital "NE" rSub { size 8{3} } T} {}
LE 1 ME 3 NE 2 T size 12{ ital "LE" rSub { size 8{1} } ital "ME" rSub { size 8{3} } ital "NE" rSub { size 8{2} } T} {}
LE 2 ME 1 NE 3 T size 12{ ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{1} } ital "NE" rSub { size 8{3} } T} {}
LE 3 ME 3 NE 1 T size 12{ ital "LE" rSub { size 8{3} } ital "ME" rSub { size 8{3} } ital "NE" rSub { size 8{1} } T} {}
LE 3 ME 2 NE 1 T size 12{ ital "LE" rSub { size 8{3} } ital "ME" rSub { size 8{2} } ital "NE" rSub { size 8{1} } T} {}
LE 3 ME 1 NE 2 T size 12{ ital "LE" rSub { size 8{3} } ital "ME" rSub { size 8{1} } ital "NE" rSub { size 8{2} } T} {}

Because the E size 12{E} {} 's are not different, there is only one arrangement LEMENET size 12{ ital "LEMENET"} {} and not six. This is true for every permutation.

Let us suppose there are n size 12{n} {} different permutations of the letters ELEMENT size 12{ ital "ELEMENT"} {} .

Then there are n 3 ! size 12{n cdot 3!} {} permutations of the letters E 1 LE 2 ME 3 NT size 12{E rSub { size 8{1} } ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{3} } ital "NT"} {} .

But we know there are 7 ! size 12{7!} {} permutations of the letters E 1 LE 2 ME 3 NT size 12{E rSub { size 8{1} } ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{3} } ital "NT"} {} .

Therefore, n 3 ! = 7 ! size 12{n cdot 3!=7!} {}

Or n = 7 ! 3 ! size 12{n= { {7!} over {3!} } } {} .

This gives us the method we are looking for.

Permutations with similar elements

The number of permutations of n size 12{n} {} elements taken n size 12{n} {} at a time, with r 1 size 12{r rSub { size 8{1} } } {} elements of one kind, r 2 size 12{r rSub { size 8{2} } } {} elements of another kind, and so on, is

n ! r 1 ! r 2 ! . . . r k ! size 12{ { {n!} over {r rSub { size 8{1} } !r rSub { size 8{2} } ! "." "." "." r rSub { size 8{k} } !} } } {}
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Find the number of different permutations of the letters of the word MISSISSIPPI.

The word MISSISSIPPI has 11 letters. If the letters were all different there would have been 11 ! size 12{"11"!} {} different permutations. But MISSISSIPPI has 4 S's, 4 I's, and 2 P's that are alike.

So the answer is 11 ! 4 ! 4 ! 2 ! size 12{ { {"11"!} over {4!4!2!} } } {}

Which equals 34,650.

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If a coin is tossed six times, how many different outcomes consisting of 4 heads and 2 tails are there?

Again, we have permutations with similar elements.

We are looking for permutations for the letters HHHHTT.

The answer is 6 ! 4 ! 2 ! = 15 size 12{ { {6!} over {4!2!} } ="15"} {} .

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In how many different ways can 4 nickels, 3 dimes, and 2 quarters be arranged in a row?

Assuming that all nickels are similar, all dimes are similar, and all quarters are similar, we have permutations with similar elements. Therefore, the answer is

9 ! 4 ! 3 ! 2 ! = 1260 size 12{ { {9!} over {4!3!2!} } ="1260"} {}
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A stock broker wants to assign 20 new clients equally to 4 of its salespeople. In how many different ways can this be done?

This means that each sales person gets 5 clients. The problem can be thought of as an ordered partitions problem. In that case, using the formula we get

20 ! 5 ! 5 ! 5 ! 5 ! = 11 , 732 , 745 , 024 size 12{ { {"20"!} over {5!5!5!5!} } ="11","732","745","024"} {}
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We summarize.

  1. Circular Permutations

    The number of permutations of n size 12{n} {} elements in a circle is

    n 1 ! size 12{ left (n - 1 right )!} {}
  2. Permutations with Similar Elements

    The number of permutations of n size 12{n} {} elements taken n size 12{n} {} at a time, with r 1 size 12{r rSub { size 8{1} } } {} elements of one kind, r 2 size 12{r rSub { size 8{2} } } {} elements of another kind, and so on, such that n = r 1 + r 2 + + r k size 12{n=r rSub { size 8{1} } +r rSub { size 8{2} } + dotsaxis +r rSub { size 8{k} } } {} is

    n ! r 1 ! r 2 ! r k ! size 12{ { {n!} over {r rSub { size 8{1} } !r rSub { size 8{2} } ! dotsaxis r rSub { size 8{k} } !} } } {}

    This is also referred to as ordered partitions .

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Combinations

Suppose we have a set of three letters A , B , C size 12{ left lbrace A,B,C right rbrace } {} , and we are asked to make two-letter word sequences. We have the following six permutations.

AB size 12{ ital "AB"} {} BA size 12{ ital "BA"} {} BC size 12{ ital "BC"} {} CB size 12{ ital "CB"} {} AC size 12{ ital "AC"} {} CA size 12{ ital "CA"} {}

Now suppose we have a group of three people A , B , C size 12{ left lbrace A,B,C right rbrace } {} as Al, Bob, and Chris, respectively, and we are asked to form committees of two people each. This time we have only three committees, namely,

AB size 12{ ital "AB"} {} BC size 12{ ital "BC"} {} AC size 12{ ital "AC"} {}

When forming committees, the order is not important, because the committee that has Al and Bob is no different than the committee that has Bob and Al. As a result, we have only three committees and not six.

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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