BUS (Discrete mathematics) Q.: There are 4 buses with full capacity of 40 people. 163 fans of local FC goes to other city with those 4 buses. Please, use proof by contradiction, that in some bus there are more than 40 people. Thank you!
 A: HINT: Suppose that Bus $1$ contains $b_1$ people, Bus $2$ contains $b_2$ people, and so on. Then we must have
$$b_1+b_2+b_3+b_4=163\;.$$
If every bus has at most $40$ people, then $b_1\le 40$, $b_2\le 40$, $b_3\le 40$, and $b_4\le 40$. Add these four inequalities; what does that tell you about $b_1+b_2+b_3+b_4$?
A: This is a simple case of the pigeonhole principle application, but the OP explicitly asked for a proof by contradiction, so, here is a proof.
Assume that there is no bus with more than 40 people in it. If $p_i$ denotes the number of the passengers in the $i$th bus, then $$b_i\leq40,\quad i=1,2,3,4$$ There are total $163$ people in all buses, therefore $$b_1+b_2+b_3+b_4=163\label{eq1}\tag{1}$$ But since $b_i\leq40$, then sum of $b_i$s would never exceed $160$, i.e., $$b_1+b_2+b_3+b_4\leq160$$ According to equation (\ref{eq1}) the sum can be replaced by $163$, so $$163\leq160$$ But this is a contradiction and thus the there is at least one bus with more than $40$ passengers.
A: Let every bus contained less than 40 peoples. Consider there are $(40-x_1)$ people in the first bus. Let there are $(40-x_2)$ people in second bus. Also there are $(40-x_3)$ in the third and $(40-x_4)$ in the fourth bus. Now from this we can easily write this     $x_1,x_2,x_3,x_4\ge 0$.
Now total number of peoples in the buses are 
$(40-x_1)+(40-x_2)+(40-x_3)+(40-x_4)=160-(x_1+x_2+x_3+x_4)$. Which is equal to 163. Then,$160-(x_1+x_2+x_3+x_4)=163$ 
$\implies (x_1+x_2+x_3+x_4)$ is a negative number( Contradiction)
A: If no bus has $40$ passengers, then we can transport at most $156=4\times39$ people.
