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.
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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