Solved

# Unique Combination Generator

Posted on 2011-04-27
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Hi Experts

I have 100 number (1..100). I want to create combinations of 10 elements per combination.  But each combination has to be unique and no numbers should meet more than once, which means ie:

Combination 1

1234567890

Combination 2
1,11,12,13,14,15,16,17,18,19,20

Is there any ways of achieving this in Excel or Access?

Your urgent help will be much appreciated.

Thanks

Anand
0
Question by:stryker11

LVL 39

Expert Comment

If you don't need random numbers, you can add 1 to some stored number. In this case combinations will be:
1,2,3,4,5,6,7,8,9,10
11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29,30 etc.
In your example I see 1 twice and 0, which is not in 1...100 (if first combination is 1,2,3,4,5,6,7,8,9,0)
0

LVL 31

Expert Comment

See attached. I created this for creating Lottery number sets but it can be used for what you want.

Populate options in top left, currently populated with 1 to 100 and 10 sets of 10.

Click "Generate Numbers"

Cheers
Rob H RNG.xls
0

Author Comment

Hi Robhenson

The real world situation is that I have 100 people attending this seminar.  We are having a meet & greet session whereby all hundred have to meet each other.  I have 10 tables and people are given 5 mins to in each table before switching tables.  There will be 10 rounds.  As such, I need to number the delegates and pre-arrange their table numbers for each round so that they will be able to meet all 100 people.
0

LVL 31

Expert Comment

For all 100 people to meet each other there are 4950 different combinations ie

person 1 must meet 99 others
person 2 must meet 98 others (having already met person 1)
person 3 must meet 97 others (having already met p1 & p2)
etc etc, the sum of which is 4950

Only 50 meetings can take place at a time, ie 1 on 1, therefore you would need 99 rounds. The fact that you have them split around ten tables is not relevant, the 50 meetings are split by havinge 5 meetings per table.

99 rounds at 5 minutes each would be 8 1/4 hours

Cheers
Rob H

0

LVL 31

Expert Comment

Apologies, ignore the timing calculation; I had five minutes per meeting whereas you have 5 mins per table which is 9 meetings therefore approx 0.5 minute per meeting so you would need 55 minutes plus changeover.

Cheers
Rob H
0

LVL 31

Accepted Solution

See attached.

The movement of people within each round will be determined by the people around the table, assume a certain element of trust among them to spend time with all people around the table.

The movement of people between tables is as follows:

P1 to P50 move up a table number eg from Table 1 to Table 2
P51 to 100 move down a table number eg from Table 10 to Table 9.

Alternatively, P1 to P50 can stay at the same tables and only P1 to P51 move, possibly a little unfair.

Cheers
Rob H Table-Combinations.xls
0

LVL 31

Expert Comment

Let mayhem commence!!!
0

LVL 31

Expert Comment

Give them all a ribbon attached to a pole in the centre and it would be like a May Pole dance no doubt occurring at numerous fairs across the country next week.

Apologies a slight typo in the alternative movement, should have been P1 to P50 stay at same tables, P51 to P100 move.

Cheers
Rob H
0

LVL 31

Expert Comment

Just done some checking and it is not complete and it may not meet your requirements anyway.

Missing example: P1 first meets P2 - P5 and P51 to P55 ie the other 9 people at Table 1; however P1 does not meet P6 to P50.

Therefore, 10 rounds won't be enough after all.

I will carry on working.

Cheers
Rob H
0

LVL 31

Expert Comment

Should have also added that by grouping the way I have, there will be groups of 5 people that stay together when moving between tables.

Cheers
Rob H
0

LVL 31

Expert Comment

45 meetings per table gives 450 per round. 4950 meetings required therefore 4950/450 = 11 rounds. Just need to figure out the combinations now. I suggest you draw up a matrix and pick out 10 at a time.
0

LVL 39

Expert Comment

I think robhenson was on a wrong way. Idea of question was clarified in comment 35482162. This is not Dating club, where you meet tet-a-tet, this is seminar, where all 10 person in one round in one table meet.
So:
1-st table 1-st round:
1 2 3 4 5 6 7 8 9 10
1-st table 2-nt round:
1 11 12 13 14 15 16 17 18 19
etc.
My opinion - question should be closed as unanswered.
0

LVL 31

Expert Comment

Dating club didn't even cross my mind!! I understood that it was a business seminar where all 100 candidates have to meet in groups of 10 at a time.

Thanks
Rob H
0

LVL 39

Expert Comment

Round 1 (table1):
1      51
2      52
3      53
4      54
5      55
Round 2 (table 2):
1      56
2      57
3      58
4      59
5      60

0

LVL 10

Expert Comment

This question has been classified as abandoned and is closed as part of the Cleanup Program. See the recommendation for more details.
0

LVL 31

Expert Comment

@modguy:

Thanks for accepting my suggestions eventhough it was not a final solution.

However, may I also point out the following:

comment 35482951 - realised error of group of 5 staying together - no response from author.
comment 35487530 - pointed out need for 11 rounds - no response from author.

Thanks
Rob H
0

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