Question 1122537
<br>
n(E) is the number of ways of choosing either none of the three $5 bills AND 3 of the four others, OR one of the three $5 bills AND two of the four others:<br>
C(3,0)*C(4,3)+C(3,1)*C(4,2) = 1*4+3*6 = 4+18 = 22.<br>
If you are just learning how to solve problems like this, notice that the ANDs indicate multiplications and the ORs indicate additions.<br>
When learning to solve problems like this, it is good practice to verify that those possibilities plus the others give the correct total of 35 total ways to choose 3 of the 7 bills.<br>
two $5 bills and one of the others: C(3,2)*C(4,1) = 3*4 = 12
three $5 bills and none of the others: C(3,3)*C(4,0) = 1*1 = 1<br>
22+12+1 = 35  CHECK!