We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways | = (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
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| = (525 + 210 + 21) | ||||||||||||
| = 756. |


Required numbers are 102, 108, 114, ... , 996
This is an A.P. in which a = 102, d = 6 and l = 996
Let the number of terms be n. Then,
a + (n - 1)d = 996
102 + (n - 1) x 6 = 996
6 x (n - 1) = 894
(n - 1) = 149
n = 150.