Question 1062364: i have a sets of number 1,2,7,7,7. what is the formula to count how many 3-digit number possible combinations using that sets, with these conditions of combinations must be unique. the result (13 combinations) are
127
172
177
217
271
277
712
717
721
727
771
772
777
And if i a have a set of 1,2,3,7,7,7, there will be 34 possible combinations.
thanks in advance.
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website! I have a sets of number 1,2,7,7,7. what is the formula to count how many
3-digit number possible combinations using that sets, with these conditions
of combinations must be unique. the result (13 combinations) are
For the number of 3-digit numbers with exactly 1 7:
Choose the position to put the 7 in 3C1=3 ways.
Choose a non-7 for the left-most remaining position in 2C1=2 ways.
Choose a non-7 for the right-most remaining position in 1C1=1 ways.
(3C1)(2C1)(1C1) = (3)(2)(1) = 6 ways.
For the number of 3-digit numbers with exactly 2 7's:
Choose the positions to put the 7's in 3C2=3 ways.
Choose a non-7 for the only remaining position in 2C1=2 ways.
(3C2)(2C1) = (3)(2) = 6 ways.
For the number of 3-digit numbers with exactly 3 7's:
3C3 = 1 way. That's 777.
Answer: 6+6+1 = 13 ways.
---------------------
And if I have a set of 1,2,3,7,7,7, there will be 34 possible combinations.
thanks in advance.
For the number of 3-digit numbers with exactly 0 (no) 7s:
Choose the position to put the 1 in 3C1 = 3 ways.
Choose the position to put the 2 in 2C1 = 2 ways.
Choose the position to put the 3 in 1C1 = 1 way.
That's (3C1)(2C1)(1C1) = (3)(2)(1) = 6
For the number of 3-digit numbers with exactly 1 7:
Choose the position to put the 7 in 3C1=3 ways.
Choose a non-7 for the left-most remaining position in 3C1=3 ways.
Choose a non-7 for the right-most remaining position in 2C1=2 ways.
That's (3C1)(3C1)(2C1) = (3)(3)(2) = 18 ways.
For the number of 3-digit numbers with exactly 2 7's:
Choose the positions to put the 7's in 3C2=3 ways.
Choose a non-7 for the only remaining position in 3C1=3 ways.
(3C2)(3C1) = (3)(3) = 9 ways.
For the number of 3-digit numbers with exactly 3 7's:
3C3 = 1 way. That's 777.
Answer: 6+18+9+1 = 34 ways.
Here they all are:
123, 127, 132, 137, 172, 173, 177, 213, 217, 231,
237, 271, 273, 277, 312, 317, 321, 327, 371, 372,
377, 712, 713, 717, 721, 723, 727, 731, 732, 737,
771, 772, 773, 777.
Edwin
|
|
|