Math Binomial theorem
posted by ann on .
I'm preparing for a test and one of the questions in my text is as follows:
In a trivia contest,Sam has drawn a topic he knows nothing about, so he makes random guesses for the 10 T/F questions. Use the binomial theorem to help find
a)the number of ways Sam can answer the test using exactly four trues
b)the number of ways that Sam can answer the test using at least one true. The answers are 210 and 1023. I have no idea how to approach this.

The binomial theorem can be applied to the 10 truefalse 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!(nr)!).
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 10241=1023. 
Thank you very much!

You're welcome!

(10/4) computed in binomial expansion? that's ten over four in parenthesis.