Answer to Question #288249 in Discrete Mathematics for helenabebe

Question

1.     There are 41 candidates approved to be allotted as the F.D.R.E. council of cabinets. Historic facts of the candidates are shown below, 8 were served as former cabinet members,14 were university presidents, 15 served as members of P.R.H., 2 served in former cabinet and as P.R.H, 4 served as university presidents and members of former cabinet, 6 were served as university presidents and members of P.R.H, 1 served in all the three positions. The find the number of candidates:

a.      Served in none of the three positions?

b.     Only university presidents?

c.      At least one of the three positions?

d.     Exactly two positions?

Answer 1

a.

$$N(C\cup Pr\cup P.R.H)=N(C)+N(Pr)+N(P.R.H)$$

$$-N(C\cap P.R.H)-N(C\cap Pr)-N(Pr\cap P.R.H)$$

$$+N(C\cap Pr\cap P.R.H)$$

$$=8+14+15-2-4-6+1=26$$

$$N((C\cup Pr\cup P.R.H)^C)=41-N(C\cup Pr\cup P.R.H)$$

$$=41-26=15$$

15 candidates Served in none of the three positions.

b.

$$N(only\ Pr)=N(Pr)-N(C\cap Pr)$$

$$-N(Pr\cap P.R.H)+N(C\cap Pr\cap P.R.H)$$

$$14-4-6+1=5$$

5 candidates served only university presidents.

c. 

$$N(C\cup Pr\cup P.R.H)= 26$$

26 candidates served at least one of the three positions.   

d.   

$$N(exactly\ two\ positiions)$$

$$=N(C\cap P.R.H)-N(C\cap Pr\cap P.R.H)$$

$$+N(C\cap Pr)-N(C\cap Pr\cap P.R.H)$$

$$+N(Pr\cap P.R.H)-N(C\cap Pr\cap P.R.H)$$

$$=2-1+4-1+6-1=9$$

9 candidates served exactly two positions.