Maths Permutations Brain Fail

[BillyBB]

I am looking for the total number of possible permutations for the following -
All answers are a six digit number, made up of four specific digits for example 1,2,3 and 4
So all answers have repetition, however ALL four digits must be used in each six digit permutation.
So first digit can be one of four and so can the second, but by the third onwards things get too complicated for me because have to factor in using all four of the digits.

[Zacs-Master]

4096

[sunny-dave]

Look at it in reverse and get all the four numbers used first

4*3*2*1

Then multiply that answer by the free choice in the last two positions

4*3*2*1*4*4 = 384

[Gizmonster]

Zacs - your answer is just 4^6 which is clearly not the answer, as Billy has said that the 6 digit number must include all of the digits 1-4.
Your answer is just a permutation of the total number of possibilities using the numbers 1-4, but doesn't necessarily include any numbers. One of these is 444444 for example, which is not allowed, so the answer in question would me lower than your answer .... I'm still trying to get my head round it .....

[Zacs-Master]

Soz

[Canary42]

4x4x4x3x2x1=384

[JJ109]

First of all you need to do this in twostages
1. 3 of a kind plus 3 singletons
There are 4 possible 3s + 3 singletons
eg 111234, 122234, 123334, 123444
But then we need the permutations for each of these 4
6! / 3! = 6 * 5 * 4 = 120
120 * 4 = 480
2 2 doubletond + 2 singletons
eg 112234, 113324,114423,223314,224413,334412
Rgen the permutions 6!/2/2 = 180
6* 180 = 1040
Total =480 + 1040 = 1520

[BillyBB]

many thanks
Brain still slightly frazzled but much better

[JJ109]

You're welcome! Apologies for my poor typing, but the electric had gone off and I was typing in the dark!

[JJ109]

Let's try and put this in better English and with a better explanation.

First of all you need to do this in two stages
1. 3 of a kind plus 3 singletons
2. 2 doubletons and 2 singletons

1. There are 4 possible 3s + 3 singletons. These are
111234, 122234, 123334, 123444
But then we need the permutations for each of these sets of 4
If all numbers had been different, then there would have been 6! = 720 permutations. BUT they are not all different. In that total of 720 you need to divide by 3! = 3 * 2 * 1 = 6 in respect of the identical permutations given by just swopping over the identical numbers.
eg for 111234 if we swop the ones for abc the permutations would consist of abc234, acb234,bac234,bca234,cab234,cba234
These 6 permutations are all identical
Therefore there are 6!/3! = 720/6 = 120 permutations for each of the 4 ways you can select a 3 of a kind plus 3 singletons
Total stage 1 = 4 * 120 = 480

Stage 2 = 2 doubletons and 2 singletons
There are 6 ways that you can select 2 doubletons and 2 singletons
112234,113324,114423,223314,224413,334412
Again if all different there would be 6! = 720 permutations for each
However we must divide by 2 for the first doubleton for the duplication of swopping one of the doubleton numbers for the other, and divide by a further two for swapping the second doubleton number for its twin.
Therefore 720/2/2 = 180
As there are 6 different ways for 2 doubletons and2 singletons and each has 180 permutations, the total permutations for 2 doubletons and 2 singletons = 180 * 6 = 1040

Adding the two stages together
Total answer = 480 + 1040 = 1520

6! / 3! = 6 * 5 * 4 = 120
120 * 4 = 480
2 2 doubletond + 2 singletons
eg 112234, 113324,114423,223314,224413,334412
Rgen the permutions 6!/2/2 = 180
6* 180 = 1040
Total =480 + 1040 = 1520

[kuiperbelt]

JJ109 /3 of a kind plus 3 singletons/

Is it your intention to keep the '3 of a kind' as 3 contiguous digits in all the possible permutations available?

[JJ109]

No it is not - that is why the permutations for each of the 4 three of a kinds and 3singletons = 6!/3! = 120

If they were kept as contiguous it would simply be 4! = 24

[ZebuSanctuary]

JJ / / If they were kept as contiguous it would simply be 4! = 24 //

Surely; 4! x 4 giving 96 permutations in the contiguous condition.

[kuiperbelt]

96 perms because the other contiguous digits are 222, 333 and 444.

[JJ109]

Yes kuiperbelt- I was only giving you the perms for 1 of the 4 scenarios

I took it for granted that you would multiply that by 4-

if you want contiguous 3 of a kind for all 4 combinations it is obviously 4 * 4! = 4 * 24 = 96

[kuiperbelt]

Thanx JJ
Neat answer btw wed 00.54 :-)

[JJ109]

Thanks :)