If I have 9 numbers (1thru 9) and I put them into three rows of three, you can mix the numbers around in any order, no numbers repeat. How many possible orders would there be?

Starting sample:
1 2 3
4 5 6
7 8 9

Thank you,
Steven

9!=9?8?7?6?5?4?3?2?1=362.880

Answer: The factorial of nine gives the number of combinations.
9!=9x8x7x6x5x4x3x2x1=362.880

362.880 different combinations of the 9 numbers are possible.
(The question marks above were unintentional autocorrect.)

There are 9 options for the first slot, 8 for the second, 7 for the third, 6 for the fourth, 5 for the fifth, 4 for the sixth, 3 for the seventh, 2 for the eighth, and 1 for the 9th. So, 9 times 8 ties 7 ... times 3, times 2 (times 1) or 9 factorial, often denoted as 9! which equals 362,880.

The 3x3 grid doesn't change anything for the calculation. In other words, you can think of it like:

123
456
789

as being the same as 123456789.

Thank you everyone for your help and input, it is much appreciated.

What I am looking for are different possible totals. Not the different permutations of a 9 digit number. That number is quite large.

I did get a math professor to look at it and his answer was 199 different possible totals of the three rows of three numbers.

Even though you can mix the numbers around you will often still get the same total. This occurrence is apparently quite frequent so the total is very low 199. I would have never guess it was that low.

Thank you again!

Steven

I apologize for misunderstanding the question. I thought you were asking for distinct 3x3 grids. Do you mind explaining what you mean by different possible totals?

Or, let me rephrase: could you provide an example of two grids that are the same total?

Let's begin with this grid:

123
456
789

Can you tell me if either (or both) of the below are considered the same total:

A) rows 2 and 3 swapped

123
789
456


or are you considering it the same total as the grid her (ie: the three numbers in row 1 are the same):

B)
321
456
789

(Or all three of these grids considered identical, as each of three rows has the same three numbers)?

I'm really curious where 199 comes from!

The more I think about it the more curious I am how this is defined. I'd love to see the math professor's explanation, because I am struggling a little I have an idea how you can get to 216, if the conditions state that:

Row 1 must have the numbers 1 2 3 (6 distinct combinations; 123, 132, 213, 231, 312, 321)

Row 2 must have the numbers 4 5 6 (6 distinct combinations; 456, 465, 546, 564, 645, 654)

Row 3 must have the numbers 7 8 9 (6 distinct combinations; 789, 798, 879, 897, 978, 987)

So it'd be 6 * 6 * 6 = 216.

I'd love to understand the actual question though, because I'm sure your math professor friend is right and I am wrong!

Hi Gang,

What Steven is referring to is the Tri-Triplet Force, something that I've explored extensively. Here are my original thoughts on it:

https://www.themagiccafe.com/forums/view......forum=82

There are indeed but 199 possible sums. They are:

[774,783,792,801,810,819,828,837,846,855,864,873,882,891,900,909,918,927,936,
945,954,963,972,981,990,999,1008,1017,1026,1035,1044,1053,1062,1071,1080,1089,1098,
1107,1116,1125,1134,1143,1152,1161,1170,1179,1188,1197,1206,1215,1224,1233,1242,1251,
1260,1269,1278,1287,1296,1305,1314,1323,1332,1341,1350,1359,1368,1377,1386,1395,1404,
1413,1422,1431,1440,1449,1458,1467,1476,1485,1494,1503,1512,1521,1530,1539,1548,1557,
1566,1575,1584,1593,1602,1611,1620,1629,1638,1647,1656,1665,1674,1683,1692,1701,1710,
1719,1728,1737,1746,1755,1764,1773,1782,1791,1800,1809,1818,1827,1836,1845,1854,1863,
1872,1881,1890,1899,1908,1917,1926,1935,1944,1953,1962,1971,1980,1989,1998,2007,2016,
2025,2034,2043,2052,2061,2070,2079,2088,2097,2106,2115,2124,2133,2142,2151,2160,2169,
2178,2187,2196,2205,2214,2223,2232,2241,2250,2259,2268,2277,2286,2295,2304,2313,2322,
2331,2340,2349,2358,2367,2376,2385,2394,2403,2412,2421,2430,2439,2448,2457,2466,2475,
2484,2493,2502,2511,2520,2529,2538,2547,2556]

Incidentally, the distribution of these 199 values and their probabilities is interesting: it's a bit ragged, but very clearly trying to be binomial.

By throwing in the digit 0, this method can be turned into either the Quin-Doublet Force or the Duo-Quintuplet force, both of which work similarly now with 10 digits total, (but with different ranges of results).

Since those early thoughts in the thread listed above, I've dug even deeper into these three forces. There's a lot more to them than meets the eye!

Thomas Henry

WOW thank you for explaining Thomas. I don't believe I have access to that forum section yet, but I'll bookmark it for later reading. I'm fascinated with mathematics + magic (currently obsessing over faro and Gilbreath-related work), so I'd love to read your thoughts on this.

Got it, Steven, you were looking for the possible sums of the 3 rows! I read "how many possible orders would there be" and immediately started thinking about distinct possibilities, not sums. My apologies for any confusion I may have added.

Hello again,

A bit more that I should have mentioned: like Steven notes, the possible sums are not spread out uniformly. So for instance, of the 362,880 ways the participant can choose and arrange the digits, the sum 1575 can occur in 3888 possible ways, while the sum of 774 can only occur in 216 ways.

If anyone's interested, I can post a matrix of all 199 possible outcomes and their probabilities.

Thomas Henry