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20P5 categorized by repeat objects

adamtj

New member
Joined
Nov 19, 2015
Messages
3
Hello, I am hoping this is in the right sub group please do let me know if I chose incorrectly.

So I have part of this solved ( I think)

The basics:
Total population of objects :20
slots to fill: 5
total population of sets: 3.2MM (20^[SUP]5[/SUP])


My goal is to organize this total population into groups.
aaaaa - 5 of a kind
aaaab - 4 of a kind, 1 single
aaabb - 3 of a kind, 1 pair
aaabc - 3 of a kind, 2 singles
aabbc - 2 pairs, 1 single
aabcd - 1 pair, 3 singles
abcde - all singles

so using [SUB]n[/SUB]P[SUB]r[/SUB] I came up with the following:

category-count
aaaaa-20
aaaab-380
aaabc -6,840
aaabb-380
aabbc -6,840
aabcd -116,280
abcde -1,860,480

This is incomplete as it sums to 1,991,220.
the next step as I was seeing it was that there are variations of each category:
Example: aaaab,aaaba,aabaa and so on.
the plan was to calculate the permutations for each and then multiplying that by the initial calculations. but this yields much too large a population. so now I am stuck.

Here is the google sheets document I have been working with:
https://docs.google.com/spreadsheets...it?usp=sharing

Any hints and help are greatly appreciated!
 

ksdhart

Full Member
Joined
Aug 1, 2014
Messages
384
You're on the right track, but you're missing many of the permutations. For instance, in the "4 of a kind, 1 single" you list only aaaab as an option. What about aaaac? Or aaaad? Similarly for "3 of a kind, 1 pair" you'd need to include aaacc, etc. When you include those, the numbers should work out right. Also, if you're only looking to have 20[SUP]5[/SUP] arrangements, then the order must not matter. aaaab is the same as aaaba because you're only interested in the fact that there are 4 a's and 1 b, not what position the b is in.
 

Ishuda

Elite Member
Joined
Jul 30, 2014
Messages
3,345
Hello, I am hoping this is in the right sub group please do let me know if I chose incorrectly.

So I have part of this solved ( I think)

The basics:
Total population of objects :20
slots to fill: 5
total population of sets: 3.2MM (20^[SUP]5[/SUP])


My goal is to organize this total population into groups.
aaaaa - 5 of a kind
aaaab - 4 of a kind, 1 single
aaabb - 3 of a kind, 1 pair
aaabc - 3 of a kind, 2 singles
aabbc - 2 pairs, 1 single
aabcd - 1 pair, 3 singles
abcde - all singles

so using [SUB]n[/SUB]P[SUB]r[/SUB] I came up with the following:

category-count
aaaaa-20
aaaab-380
aaabc -6,840
aaabb-380
aabbc -6,840
aabcd -116,280
abcde -1,860,480

This is incomplete as it sums to 1,991,220.
the next step as I was seeing it was that there are variations of each category:
Example: aaaab,aaaba,aabaa and so on.
the plan was to calculate the permutations for each and then multiplying that by the initial calculations. but this yields much too large a population. so now I am stuck.

Here is the google sheets document I have been working with:
https://docs.google.com/spreadsheets...it?usp=sharing

Any hints and help are greatly appreciated!
I don't quite understand. I think what you are saying is you have a set of 20 distinct objects, i.e. something like
S={o[SUB]j[/SUB], j=1, 2, 3, ..., 20; o[SUB]i[/SUB]\(\displaystyle \ne\)o[SUB]j[/SUB] if i\(\displaystyle \ne\)j}
that is, for example,
S={a, b, c, ..., r, s, t}
and you want to form a group of sets with particular properties. Assuming this is true, first let
T = {1, 2, 3, ..., 20}
so I won't have to write out the full number set each time.

Now the groups (set of sets) you want are
s[SUB]1[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB]}, j \(\displaystyle \epsilon\)T} [thus 20 sets of type {a, a, a, a, a}]
s[SUB]2[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB]}, j \(\displaystyle \epsilon\)T; j \(\displaystyle \epsilon\)T-{j}} [4 of a kind]
s[SUB]3[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB], o[SUB]i[/SUB]}, j \(\displaystyle \epsilon\)T; j \(\displaystyle \epsilon\)T-{j}} [Full House]
etc.
In each of these sets, order doesn't matter. That is, for example, in s[SUB]2[/SUB] {a, a, a, b, b} would be considered the same as {b, a, a, a, b}

A page I recommend quite a bit for permutations and combinations is
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
The whole page makes for an interesting read IMO but the particular section for this problem is "Combinations with Repetition"
 

adamtj

New member
Joined
Nov 19, 2015
Messages
3
You're on the right track, but you're missing many of the permutations. For instance, in the "4 of a kind, 1 single" you list only aaaab as an option. What about aaaac? Or aaaad? Similarly for "3 of a kind, 1 pair" you'd need to include aaacc, etc. When you include those, the numbers should work out right. Also, if you're only looking to have 20[SUP]5[/SUP] arrangements, then the order must not matter. aaaab is the same as aaaba because you're only interested in the fact that there are 4 a's and 1 b, not what position the b is in.
Hello thank you for responding.

Two things;
One in my situation order does matter.
Two, the aaaab was not meant to represent that there are 4 of one kind one of the other.

so for example, given 20 objects and 5 slots in the form aaaab there are 380 sets. but there are 5 ways to organize a 4 of a kind plus one single. [5!/(4!*1)]
aaaab
aaaba
aabaa
abaaa
baaaa
so a total of 380*5=1900 sets.
the problem is that when you get the 1 pair and three single. doing [5!/(2!*1*1!*1!)]=60 doesn't work.
116,280*60=6976800 as that is larger then the entire population which is 3.2M
 

adamtj

New member
Joined
Nov 19, 2015
Messages
3
I don't quite understand. I think what you are saying is you have a set of 20 distinct objects, i.e. something like
S={o[SUB]j[/SUB], j=1, 2, 3, ..., 20; o[SUB]i[/SUB]\(\displaystyle \ne\)o[SUB]j[/SUB] if i\(\displaystyle \ne\)j}
that is, for example,
S={a, b, c, ..., r, s, t}
and you want to form a group of sets with particular properties. Assuming this is true, first let
T = {1, 2, 3, ..., 20}
so I won't have to write out the full number set each time.

Now the groups (set of sets) you want are
s[SUB]1[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB]}, j \(\displaystyle \epsilon\)T} [thus 20 sets of type {a, a, a, a, a}]
s[SUB]2[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB]}, j \(\displaystyle \epsilon\)T; j \(\displaystyle \epsilon\)T-{j}} [4 of a kind]
s[SUB]3[/SUB] = {{o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]j[/SUB], o[SUB]i[/SUB], o[SUB]i[/SUB]}, j \(\displaystyle \epsilon\)T; j \(\displaystyle \epsilon\)T-{j}} [Full House]
etc.
In each of these sets, order doesn't matter. That is, for example, in s[SUB]2[/SUB] {a, a, a, b, b} would be considered the same as {b, a, a, a, b}

A page I recommend quite a bit for permutations and combinations is
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
The whole page makes for an interesting read IMO but the particular section for this problem is "Combinations with Repetition"
Hello Ishuda,

Thank you for your reply, you are 100% correct other then that in my situation order does matter. I require the sum off all variations with the different sets ie aaaab,+aaaba, ect.


 
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