<p>These are challenging..well for me...combinatorics problems! Please help me if possible.
1.) How many subsets of six intergers chosen (without repetition) from 1,2, ..., 20 (1-20) are there with no consecutive integers (e.g.if 5 is in the subset then 4 and 6 cannot be in it but if 1 is in the set only 2 cannot be in it)?
2.) How many 90digit numbers are there with twice as many different odd digits involved as different even digits (e.g. 945222123 with 9, 3,5,1 odd and 2,4 even)?</p>
<p>Your help is greatly appreciated..if possible an explanation also ! Thanks</p>
<p>1) 20<em>18</em>16<em>14</em>12<em>10 i think...?
2) 2 cases : # of odd digit is either 2 or 4. (My guess is 0 odd digit is not on this.) For case1 there is only one even, giving 1+2=3 possible choice for each digit. 3^90</em>10<em>5 (10=C(5,2), 5=C(5,1)) For case2, 6 possible choice. 6^90</em>10<em>5. But the cases with 0 in front are not covered. So... 3^90</em>10<em>5 - 3^89</em>10 + 6^90<em>10</em>5 - 6^89<em>4</em>5</p>
<p>I thought it was going to something like SAT prep q's...
Since they aren't, I'm sure I got this wrong, but I'm posting it so that my time wasn't for nothing</p>