The binomial theorem can be applied to the 10 true-false questions as follows:
Let (T+F)^10 represent the choices Sam could make, i.e. for each question, he enters either true or false.
We know that:
(T+F)^10=
T^10+10*F*T^9+45*F^2*T^8+120*F^3*T^7+210*F^4*T^6+252*F^5*T^5+210*F^6*T^4+120*F^7*T^3+45*F^8*T^2+10*F^9*
T+F^10
The coefficients tell us the number of ways the combinations of true/false can be made.
For example:
There is only one way to write 10 trues (obviously).
But there are 10 ways to write 1 false and 9 trues (a false at each of the 10 questions), and so on.
The coefficients are obtained, fortunately, easily by the formula for "n choose r", or C(n,r), or n!/(r!(n-r)!).
So for 4 trues and 6 falses, the number of ways is C(10,4)=10*9*8*7/(1*2*3*4)=210.
Similarly, to have at least one true, we can do a summation of C(10,1),C(10,2)....C(10,10) to get 1023.
Alternatively, we note that there are 2^10=1024 ways to do the exam, out of which only one case (10 false) does not have at least one true. So the number of ways for at least one true is 1024-1=1023.
