MATH 504 - Final Exam (Exercise 5.3 To Chapter Exercise)
MATH 504 - Final Exam (Exercise 5.3 To Chapter Exercise)
MATH 504 - Final Exam (Exercise 5.3 To Chapter Exercise)
∴There are 120 ways of a committee can be formed from a group of 10 members.
b) How many triangles can be drawn using the vertices chosen from these points?
Solution:
n=10, r=3
n! 10 ! 10 ! 10 × 9 ×8 720
C =
10 3 = = = = =120
r ! ( n−r ) ! 3! (10−3 ) ! 3! 7 ! 3 ×2 6
9. At MSA Academy Inc., a four-person committee is chosen from 15 teachers and a principal. In how many ways
can the committee be formed if
a) The principal must be in the committee?
Solution: n=16, r=4
n! 16 ! 16 ! 16 ×15 ×14 × 13 43680
C = = = = = =1820 ways
r ! ( n−r ) ! 4 ! ( 16−4 ) ! 4 ! 12 ! 4 × 3 ×2 24
16 4
15. Suppose 6 female and 5 male applicants have been successfully screened for 5 positions. In how many ways can
the following compositions be selected?
a) 5 people regardless of sex
Solution:
11 ! 11! 11×10 ×9 × 8× 7 55440
C= = = = =462 ways
11 5
5! (11−5 ) ! 5 !6! 5 × 4 ×3 × 2 120
b) 3 females and 2 males
Solution:
C × 5C2 = 20×10=¿200 ways
6 3
c) At least 3 females
Solution:
6C3 × 5C2 = 20×10=¿200 ways
d) 4 females and 1 male
Solution:
C × 5C1 = 15 ×1= 15×5 =75 ways
6 4
e) 5 females
Solution:
6! 6!
C = = =6 ways
5! (6−5 ) ! 5 !
6 5
16. How many 4-person committee are possible from a group of 10 people including Francis and Jojo if ;
a) There are no restrictions?
Solution:
12! 12 ×11 ×10 × 9 11800
C = = = =495 ways
12 4
4 ! ( 12−4 ) ! 4 ×3 × 2 24
b) Both Francis and Jojo must be in the committee?
Solution:
12! 12 ×11 ×10 × 9 11800
C = = = =495 ways
12 4
4 ! ( 12−4 ) ! 4 ×3 × 2 24
c) Either Francis or Jojo (but not both) must be in the committee?
Solution:
11 ! 11 ×10 ×9 × 8 7920
C = = = =330 ways
11 4
4 ! ( 11−4 ) ! 4 ×3 ×2 24
17. A warehouse receives a shipment of 20 computers, of which five are defective. Eight computers are then
randomly selected and then delivered to a store.
a) In how many ways can the store receives no defective computer?
15 ! 15! 15 ×14 × 13× 12× 11×10 ×9
C= = = =6435 ways
8 ! ( 15−8 ) ! 8 ! 7 ! 7 × 6 ×5 × 4 ×3 ×2
15 8
b) In how many ways can the store receive three defective computer?
15C5×5C3=3003 ×10= 30, 030 ways
c) In how many ways can the store receive all five defective computer?
15C3×5C5= 455 ×1= 455 ways
18. Find the number of different poker hands that contains exactly 3 kings, while the other remaining 2 cards do not
form a pair.
4C3 × 13C2 = 2 × 78 = 156 ways
19. Find the number of full houses in a poker hand, that is, the number of poker hands with three of a kind and 2 of
a kind.