Why the factorial of 0 is 1
Simplification of factorial finding
Using this new information it is easy to find the factorial of larger numbers. Here are a few things to remember.
- A set of objects that are placed in a certain order or arrangement will have more gaps than the number of objects by one.One object has two gaps that could allow an additional object. Two objects will have three gaps that can be filled by a third object. The same goes for large numbers. Twenty objects will have twenty one different places where a new object can be added.
- Keeping that in mind, it appears if we already know the factorial of N objects (say N!), the factorial of N+1 objects is simply the product of N+1 and the factorial of N.
(N+1)! = (N+1)•N!
That formula seems to work correctly with finding the factorial of three when the factorial of two is known. Since,
2! = 2
it should follow that
3! = 3(2!) = 3(2) = 6
which is true as we’ve discovered just now.
We can calculate the factorials of 4 to 10 similarly.
4! = 4(3!) = 4(6) = 24
5! = 5(4!) = 5(24) = 120
6! = 6(5!) = 6(120) = 720
7! = 7(6!) = 7(720) = 5040
8! = 8(7!) = 8(5040) = 40320
9! = 9(8!) = 9(40320) = 362880
10! = 10(9!) = 10(362880) = 3628800
Needless to say, factorials get too large too soon.
Factorial of four
Let’s see if the simplified method for finding factorials works for four, which by the looks of it seems only 20% as hard as the factorial of five.
Arrangement | Count | Arrangement | Count | |||||||
---|---|---|---|---|---|---|---|---|---|---|
1st | 2nd | 3rd | 4th | 1st | 2nd | 3rd | 4th | |||
A | B | C | D | 1 | B | A | C | D | 13 | |
A | B | D | C | 2 | B | A | D | C | 14 | |
A | D | B | C | 3 | B | D | A | C | 15 | |
D | A | B | C | 4 | D | B | A | C | 16 | |
A | C | B | D | 5 | B | C | A | D | 17 | |
A | C | D | B | 6 | B | C | D | A | 18 | |
A | D | C | B | 7 | B | D | C | A | 19 | |
D | A | C | B | 8 | D | B | C | A | 20 | |
C | A | B | D | 9 | C | B | A | D | 21 | |
C | A | D | B | 10 | C | B | D | A | 22 | |
C | D | A | B | 11 | C | D | B | A | 23 | |
D | C | A | B | 12 | D | C | B | A | 24 |
Those were the twenty four unique ways one could arrange four objects.
Factorial of five
To arrange five objects one would have to place four objects in any one of those orders and the fifth in any of the five available places. Doing that for every single order would yield a total of 120 ways in which five objects could be arranged. Let’s use a simpler table here.
Base | Arrangements | |||||
---|---|---|---|---|---|---|
1st | 2nd | 3rd | 4th | 5th | ||
ABCD | ABCDE | ABCED | ABECD | AEBCD | EABCD | |
ABDC | ABDCE | ABDEC | ABEDC | AEBDC | EABDC | |
ADBC | ADBCE | ADBEC | ADEBC | AEDBC | EADBC | |
DABC | DABCE | DABEC | DAEBC | DEABC | EDABC | |
ACBD | ACBDE | ACBED | ACEBD | AECBD | EACBD | |
ACDB | ACDBE | ACDEB | ACEDB | AECDB | EACDB | |
ADCB | ADCBE | ADCEB | ADECB | AEDCB | EADCB | |
DACB | DACBE | DACEB | DAECB | DEACB | EDACB | |
CABD | CABDE | CABED | CAEBD | CEABD | ECABD | |
CADB | CADBE | CADEB | CAEDB | CEADB | ECADB | |
CDAB | CDABE | CDAEB | CDEAB | CEDAB | ECDAB | |
DCAB | DCABE | DCAEB | DCEAB | DECAB | EDCAB | |
BACD | BACDE | BACED | BAECD | BEACD | EBACD | |
BADC | BADCE | BADEC | BAEDC | BEADC | EBADC | |
BDAC | BDACE | BDAEC | BDEAC | BEDAC | EBDAC | |
DBAC | DBACE | DBAEC | DBEAC | DEBAC | EDBAC | |
BCAD | BCADE | BCAED | BCEAD | BECAD | EBCAD | |
BCDA | BCDAE | BCDEA | BCEDA | BECDA | EBCDA | |
BDCA | BDCAE | BDCEA | BDECA | BEDCA | EBDCA | |
DBCA | DBCAE | DBCEA | DBECA | DEBCA | EDBCA | |
CBAD | CBADE | CBAED | CBEAD | CEBAD | ECBAD | |
CBDA | CBDAE | CBDEA | CBEDA | CEBDA | ECBDA | |
CDBA | CDBAE | CDBEA | CDEBA | CEDBA | ECDBA | |
DCBA | DCBAE | DCBEA | DCEBA | DECBA | EDCBA |
This table clearly shows 120 entries to the number of arrangements, which should be obvious if you know your tables of 5 or that of 24.