Can anyone explain a perm

to explain rather than doing the counting for you, you have these bets:

Doubles:

AB
AC
AD
AE
BC
BD
BE
CD
CE
DE

Trebles

ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE

Four-folds

ABCD
ABCE
ABDE
ACDE
BCDE

Five-folds

ABCDE

So for two winners, assume A & B win and count your winning lines, for three assume A, B & C, for four assume A, B, C & D and clearly for five A, B, C, D & E

Simple approach for any perm you wish to do
Say you do trebles from 6 selections
so you take the second number which is six selections & the do multiplication of

6X5X4= 120 ( 3 multiplications for trebles )

Then take the first part trebles ( 3 selections ) & do

3x2x1 =6

Then divide the second total into the first 120 divided by 6 =20 selections.

Same approach for 4 folds from 6 except you times 4 numbers for four folds.

so

6x5x4x3 = 360

4x3x2x1 =24

24 into 360 = 15

this can be done with say 8 folds from 12 as well as any other perm you wish do do.

Very simple fraction for working out `how many from how many`, similar to pumphol`s

Start with the number of selections, in this case I`ll pick `7`
So, write down `7` as the first entry on the top half of the fraction; then count down all the numbers, down to 1, giving us 7 6 5 4 3 2 1, which is the top half of the fraction.
Now, let`s assume you want to know `how many five-timers from seven selections?`; so start the bottom line of the fraction with that `5`, and count down to `1` again, giving us

7 6 5 4 3 2 1
-------------
5 4 3 2 1


Next, subtract your `how many` number, (5) from your `total selections` (7), leaving in this case `2`. Add this number to the bottom half of the fraction, and count down to one again.

7 6 5 4 3 2 1
-------------
5 4 3 2 1 2 1

Cancel out any numbers that are in both top and bottom, leaving

7 6
---
2 1

7x6=42
------
2x1-2

42
--
2  = 21 five-timers from seven selections



Another example; how many four-timers from ten selections?

10 9 8 7 6 5 4 3 2 1               
--------------------
4 3 2 1 6 5 4 3 2 1 

which becomes 

10x9x8x7
--------
4x3x2x1

but we can multiply the bottom `4` and `2` to make `8`, which then cancels the top half `8`, so

10x9x7
------
3x1

Cancel the bottom `3` with the top `9`, leaving

10x3x7
------
1x1 

210
---
1

so 210 four-timers from 10 selections



Going back to the OP, `doubles from five selections`

5 4 3 2 1
---------
2 1 3 2 1

5x4
----
2x1  =10 doubles


5 4 3 2 1
---------
3 2 1 2 1

5x4
---
2x1= 10 trebles


5 4 3 2 1
---------
4 3 2 1 1

5
-
1  = 5 four-timers

Here's a web page that will do  it for you:

https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php

Long, long,time  ago when I was a betting shop manager,I had a punter who used to do every day combinations in the same races,
eg 3x2x4x3 but I'm b.ggared if I can remember how to work it out.

Simple check: if you know one, you know one other: EG 7 selections
Singles = 6somes (7)
Doubles = 5somes (21)
Trebles - 4somes (35)

pascals triangle add above numbers to produce new number....


row 1....................................1.......................................................1 selection, 1 single
row 2 ..............................1......2........1..............................................2 selections , 2 singles, 1 double
row 3........................1...,,,,,,3.......3........1........................................3 selections 3 singles, 3 doubles, 1 treble
row 4 ........."..........1.......4.......6.........4..........1..............................4 selections , 4 singlles, 6 doubles, 4 trebles, 1 fourtimer
row 5 ................1.......5......10......10........5...........1.....................etc
row 6 ...........1........6.......15.....20......15.....,,, 6.........1................etc
row 7.......1.......7.........21.....35.....35.......21........7........1.......dunter1 bet......7 selections , 7 singles, 21 doubles ...etc

etc

Keegan