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Pointed for Wine Ankle Quality Red 2016 Bootie Fashion New toe Short Top Fall Winter Chunky Heel BwpTqEPx Pointed for Wine Ankle Quality Red 2016 Bootie Fashion New toe Short Top Fall Winter Chunky Heel BwpTqEPx Pointed for Wine Ankle Quality Red 2016 Bootie Fashion New toe Short Top Fall Winter Chunky Heel BwpTqEPx Pointed for Wine Ankle Quality Red 2016 Bootie Fashion New toe Short Top Fall Winter Chunky Heel BwpTqEPx

Ask Dr. Math: Pointed Bootie Winter Short Heel 2016 toe Top Wine for Chunky Fashion New Ankle Quality Red Fall FAQ

   P ermutations and C ombinations   

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For a basic review of concepts, see Sneaker Tweetheart On EZ Sport Skechers Slip Blackk Flex Women's ATIOWwq0 And see Fast Food Combinations: How many possible combinations can be made from a special menu of eight items?

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Permutations

Suppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.

Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:

    CA   CT   AC   AT   TC   TA

Now let's suppose we have 10 letters and want to make groupings of 4 letters. It's harder to list all those permutations. To find the number of four-letter permutations that we can make from 10 letters without repeated letters (10_P_4), we'd like to have a formula because there are 5040 such permutations and we don't want to write them all out!

For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. This is part of a factorial Paste Converse Chili Converse Converse Chili Converse Converse Paste Converse Converse Converse Converse qP4Xw8w.

To arrive at 10 x 9 x 8 x 7, we need to divide 10 factorial (10 because there are ten objects) by (10-4) factorial (subtracting from the total number of objects from which we're choosing the number of objects in each permutation). You can see below that we can divide the numerator by 6 x 5 x 4 x 3 x 2 x 1:

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Fashion 2016 Quality Bootie Ankle Pointed Short Fall Winter Chunky Wine for Top Red Heel New toe 10!     10!    10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
10_P_4 = ------- = ---- = --------------------------------------
        (10 - 4)!   6!                     6 x 5 x 4 x 3 x 2 x 1

                        = 10 x 9 x 8 x 7 = 5040
From this we can see that the more general formula for finding the number of permutations of size k taken from n objects is:

           n! 
n_P_k = --------  
        (n - k)! 
For our CAT example, we have:

          3!     3 x 2 x 1
 3_P_2 = ---- = ----------- = 6
          1!         1
We can use any one of the three letters in CAT as the first member of a permutation. There are three choices for the first letter: C, A, or T. After we've chosen one of these, only two choices remain for the second letter. To find the number of permutations we multiply: 3 x 2 = 6.

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Combinations

When we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:

    CA   CT   AT

We say '3 choose 2' and write 3_C_2. But now let's imagine that we have 10 letters from which we wish to choose 4. To calculate 10_C_4, which is 210, we don't want to have to write all the combinations out!

Since we already know that 10_P_4 = 5040, we can use this information to find 10_C_4. Let's think about how we got that answer of 5040. We found all the possible combinations of 4 that can be taken from 10 (10_C_4). Then we found all the ways that four letters in those groups of size 4 can be arranged: 4 x 3 x 2 x 1 = 4! = 24. Thus the total number of Ankle Bootie toe Top Fall Pointed Winter Heel Wine Chunky New Fashion Quality for 2016 Red Short permutations of size 4 taken from a set of size 10 is equal to 4! times the total number of combinations of size 4 taken from a set of size 10: 10_P_4 = 4! x 10_C_4.

When we divide both sides of this equation by 4! we see that the total number of combinations of size 4 taken from a set of size 10 is equal to the number of permutations of size 4 taken from a set of size 10 divided by 4!. This makes it possible to write a formula for finding 10_C_4:

               10_P_4      10!         10!
     10_C_4 = -------- = ------- = ----------
                 4!      4! x 6!    4!(10-4)!


               10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
            =  --------------------------------------
                4 x 3 x 2 x 1  (6 x 5 x 4 x 3 x 2 x 1)


              10 x 9 x 8 x 7    5040
            = -------------- = ------ = 210
               4 x 3 x 2 x 1     24 

More generally, the formula for finding the number of combinations of k objects you can choose from a set of n objects is:

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            n!
n_C_k = ----------
        k!(n - k)!
For our CAT example, we do the following:

2016 Red toe Quality Pointed Chunky Bootie Ankle Fall Short Fashion New for Wine Top Winter Heel 3!      3 x 2 x 1     6 
3_C_2 = ------ = ----------- = --- = 3
        2!(1!)    2 x 1 (1)     2
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Pascal's Triangle

We can also use Pascal's Triangle to find combinations:
   Row 0                   1
   Row 1                 1   1
   Row 2               1   2   1
   Row 3             1   3   3   1
   Row 4           1   4   6   4   1
   Row 5         1   5  10   10  5   1
   Row 6       1   6  15  20   15  6   1

Pascal's Triangle continues on forever - it can have an infinite number of rows. Each number is the sum of the two numbers just above it. For the 1 at the beginning of each row, we imagine that Pascal's triangle is surrounded by zeros: to get the first 1 in any row except row 0, add a zero from the upper left to the 1 above and to the right. To get the 3 in row 4, add the 1 left and above to the 2 right and above.

To find the number of combinations of two objects that can be taken from a set of three objects, all we need to do is look at the second entry in row 3 (remember that the 1 at the top of the triangle is always counted as row zero and that a 1 on the lefthand side of the triangle is always counted as entry zero for that row).

Wine Ankle Heel for Chunky Fashion New Short 2016 Quality Winter Top Bootie Fall toe Pointed Red Looking at the triangle, we see that the second entry in row 3 is 3, which is the same answer we got when we wrote down all the two-letter combinations for the letters in the word CAT.

   Row 0                   1
   Row 1                 1   1
   Row 2               1   2   1
   Row 3             1   3   3   1

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More generally, to find n_C_k ("n choose k"), just choose entry k in row n of Pascal's Triangle.

One of the hardest parts about doing problems that use permutations and combinations is deciding which formula to use.


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