Can sci solve a basic combinatorics problem?

>>8449026
4*7*7

>>8449026
4*6*6

(4*6)+(4*6*5)

Since it is a three digit number it cant start with zero. Since it is odd it cant end with an even number. therefore the zero must be the middle number.

Thus choose one of the four odd numbers to be last and then one of the remaining six to be first.

6*4 = 24.

For super-brainlets:

There are 144 numbers whose digits are distinct integers from the set {0, 1, 2, 3, 4, 5, 6, 7}

Solution:
1) Four possible values for the last digit (1, 3, 5, 7) as the last digit of any odd number must also be odd.
2) 7 possible values for the mid. digit. However, whether zero is used as the digit affects the pool of digits for the remaining init. digit, as explained in step 3.
3) 0 cannot be used as the initial digit, as it would break the 3-digit requirement. The "pool" of values for the init. digit is reduced by 1. If 0 is chosen as the mid. digit, this means the pool is only reduced by 1 (as 0 has already been removed, nothing else we picked for the middle digit). However, if 0 is not chosen, it is reduced by 2 (0 and that non-0 value).
3) To calculate the no. of numbers satisfying the requirements without 0 as a mid. digit, we multiply 4 (odds) by 6 (full pool - prev. val. and 0) and 5 (full pool - 0 and 2 prev. val) to get 120.
4) To calculate the no. of numbers satisfying the requirements with 0 as a mid digit, we multiply 4 (initial pool of odd values) by 1 (always 0) by 6 (initial pool minus chosen value and 0) to get 24. The total number of values is 120 + 24, or 144.

(The numbers in question are 103, 105, 107, 123, 125, 127, 135, 137, 143, 145, 147, 153, 157, 163, 165, 167, 173, 175, 201, 203, 205, 207, 213, 215, 217, 231, 235, 237, 241, 243, 245, 247, 251, 253, 257, 261, 263, 265, 267, 271, 273, 275, 301, 305, 307, 315, 317, 321, 325, 327, 341, 345, 347, 351, 357, 361, 365, 367, 371, 375, 401, 403, 405, 407, 413, 415, 417, 421, 423, 425, 427, 431, 435, 437, 451, 453, 457, 461, 463, 465, 467, 471, 473, 475, 501, 503, 507, 513, 517, 521, 523, 527, 531, 537, 541, 543, 547, 561, 563, 567, 571, 573, 601, 603, 605, 607, 613, 615, 617, 621, 623, 625, 627, 631, 635, 637, 641, 643, 645, 647, 651, 653, 657, 671, 673, 675, 701, 703, 705, 713, 715, 721, 723, 725, 731, 735, 741, 743, 745, 751, 753, 761, 763, and 765.)

File: autism impossible.jpg (55KB, 567x562px) Image search: [Google]

55KB, 567x562px

>>8449159
>(The numbers in question are 103, 105, 107, 123, 125, 127, 135, 137, 143, 145, 147, 153, 157, 163, 165, 167, 173, 175, 201, 203, 205, 207, 213, 215, 217, 231, 235, 237, 241, 243, 245, 247, 251, 253, 257, 261, 263, 265, 267, 271, 273, 275, 301, 305, 307, 315, 317, 321, 325, 327, 341, 345, 347, 351, 357, 361, 365, 367, 371, 375, 401, 403, 405, 407, 413, 415, 417, 421, 423, 425, 427, 431, 435, 437, 451, 453, 457, 461, 463, 465, 467, 471, 473, 475, 501, 503, 507, 513, 517, 521, 523, 527, 531, 537, 541, 543, 547, 561, 563, 567, 571, 573, 601, 603, 605, 607, 613, 615, 617, 621, 623, 625, 627, 631, 635, 637, 641, 643, 645, 647, 651, 653, 657, 671, 673, 675, 701, 703, 705, 713, 715, 721, 723, 725, 731, 735, 741, 743, 745, 751, 753, 761, 763, and 765.)

>>8449159
I think >>8449063
would have sufficed

>>8449026
4*7*6

4 valid unit digits then a choice of 7 from the remaining digits for the 10s digit, then a choice of 6 from the remaining digits for the 100s digit

>>8449159
Doing the lords work