how many different ways can you write the following using
probability, permutations or combinations
a4b2c5d
There are four factors a4, b2, c5, and d.
There are 4 choices of factors to go first.
Then for each of those 4 possible ways to choose the
first factor, there are 3 factors remaining to choose
for the 2nd factor. That's 4×3 or 12 ways to choose
the first two factors.
Then for each of those 4×3 or 12 possible ways to choose
the first two factors, there are 2 factors remaining to
choose for the 3rd factor. That's 4×3×2 or 24 ways to choose
the first three factors.
Then for each of those 4×3×2 or 24 possible ways to choose
the first two factors, there is only 1 factor remaining to
choose for the 4th factor. That's 4×3×2×1, or still 24 ways
to choose the first four factors.
It can also be called 4! or "the number of permutations of 4
things taken 4 at a time.
These are
1. a4b2c5d
2. a4b2dc5
3. a4c5b2d
4. a4c5db2
5. a4db2c5
6. a4dc5b2
7. b2a4c5d
8. b2a4dc5
9. b2c5a4d
10. b2c5da4
11. b2da4c5
12. b2dc5a4
13. c5a4b2d
14. c5a4db2
15. c5b2a4d
16. c5b2da4
17. c5da4b2
18. c5db2a4
19. da4b2c5
20. da4c5b2
21. db2a4c5
22. db2c5a4
23. dc5a4b2
24. dc5b2a4
Edwin
AnlytcPhil@aol.com
