SOLUTION: List all the permutations of the 4 objects a, b, c, d, and e, taken 3 at a time.

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Question 1191406: List all the permutations of the 4 objects a, b, c, d, and e, taken 3 at a time.
Found 2 solutions by Theo, math_tutor2020:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the permutation formula is:
p(n,x) = n! / (n-x)!
when n = 4 and x = 3, the formula becomes:
p(4,3) = 4! / (4-3)! = 4! / 1! = (4*3*2*1!) / 1! = (4*3*2) = 24
those permutations would be:
abc acb bac bca cab cba
abd adb bad bda dab dba
acd adc cad cda dac dca
bcd bdc cbd cdb dbc dcb


Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Your teacher made a typo. There are 5 objects in the set {a,b,c,d,e} and not four items.

There are 5 items to pick from and 3 slots to fill.
This yields 5*4*3 = 20*3 = 60 permutations.

You can use the nPr formula with n = 5 and r = 3 as an alternative route.


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Here's a systematic way to list all the permutations.
Naturally we can start with the letter 'a'. Then followed by b and c
abc

Next, replace the last item c with d
abd

Repeat but now replace with e
abe

So far we have these 3 permutations that start with "ab"
abc
abd
abe
So we have the "ab" fixed and the last slot goes c,d,e in that order.
I find it's easiest to stick to alphabetical order.

Now onto the next batch.
We'll keep 'a' the first letter, but we'll use c as the second slot this time
acb
acd
ace
The "ac" is fixed and we have b,d,e in alphabetical order in the third slot.

Now try d in the second slot
adb
adc
ade

Lastly, we'll have 'e' in the second slot
aeb
aec
aed

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To recap so far, we have these 12 permutations with 'a' as the first letter
  1. abc
  2. abd
  3. abe
  4. acb
  5. acd
  6. ace
  7. adb
  8. adc
  9. ade
  10. aeb
  11. aec
  12. aed
That covers every possible permutation where 'a' is listed first.

Here's every permutation where 'b' is listed first
Take note that we have blocks of 3 when considering the second slot being fixed.
I'll color-code those blocks in alternating colors of blue and red
  1. bac
  2. bad
  3. bae
  4. bca
  5. bcd
  6. bce
  7. bda
  8. bdc
  9. bde
  10. bea
  11. bec
  12. bed
Like with the stuff that started with 'a', we have 12 permutations here.
Each of the other letters will also have 12 permutations each (so that helps show why the 5 groups of 12 each yields 5*12 = 60 permutations in all).

I'll let you determine the rest.
Here's a really useful calculator to not only calculate combinations/permutations, but it also lists out the actual items (and not just a single number).
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
You'll have to scroll down and click "list" after filling in the appropriate items of course. Make sure to have the "repetition is allowed?" set to "no".