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WORDED PROBLEMS

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ABIGAIL Y. DUYOR
BSE-ENG 3A
PERMUTATION
1. In how many ways can 5 people arrange themselves in a row for picture taking?
Solution: n=5
r=5
P(5,5)= 5x4x3x2x1
=120 possible pictures
2. Find the number of permutations of the letters of the word STATISTICS.
Solution: There are 10 letters of the word. Assuming that the letters are distinct, there are
P(10, 10) = 10! permutations.
However, we have to take into consideration that the 3 S’s are alike, the 3 T’s are alike, and the 2 I’s
are alike. The permutations of the 3 S’s is P(3, 3) = 3!. The permutations of the 3 T’s is P(3, 3) = 3!.
The permutation of the 2 T’s is P(2, 2) = 2!
So, we must divide 10! by 3! 3! 2! in order to eliminate the duplicates.
Thus,
P=
10!
3!3!2!
=
10๐‘ฅ9๐‘ฅ8๐‘ฅ7๐‘ฅ6๐‘ฅ5๐‘ฅ4๐‘ฅ3๐‘ฅ2๐‘ฅ1
3๐‘ฅ2๐‘ฅ1 3๐‘ฅ2๐‘ฅ1 2๐‘ฅ1
=
3,628,800
72
= 50, 400 permutations
COMBINATIONS
1. How many different sets of 5 cards each can be formed from a standard deck of 52 cards?
Solution: n=52
r=5
C(n, r) =
๐‘›!
๐‘Ÿ!(๐‘›−๐‘Ÿ)!
=
52!
5!(52−5)!
=
52!
5!47!
=
52๐‘ฅ51๐‘ฅ50๐‘ฅ49๐‘ฅ48
5๐‘ฅ4๐‘ฅ3๐‘ฅ2๐‘ฅ1
=
311,875,200
120
(Cancel the 47! in the 52!) and
cancel the 47! in the denominator.)
sets of cards
= 2,598,960
2. In a 10-item Mathematics problem-solving test, how many ways can you select 5 problems to solve?
Solution: n=10
r=5
ABIGAIL Y. DUYOR
BSE-ENG 3A
๐‘›!
๐‘Ÿ!(๐‘›−๐‘Ÿ)!
C(n, r)=
=
10!
5!(10−5)!
3 2
=
10๐‘ฅ9๐‘ฅ8๐‘ฅ7๐‘ฅ6๐‘ฅ5๐‘ฅ4๐‘ฅ3๐‘ฅ2๐‘ฅ1
5๐‘ฅ4๐‘ฅ3๐‘ฅ2๐‘ฅ1
= 3x2x7x6
= 252 ways
PROBABILITY
1. A survey was conducted to determine girl’s favorite brand of clothes. Each girl chose only one
brand from the list of brands Uniqlo, H&M, Shein, Zara or Penshoppe. What is the probability
that a girl’s favorite brand is Shein?
BRANDS OF
CLOTHES
NUMBER OF
GIRLS
UNIQLO
H&M
SHEIN
ZARA
PENSHOPPE
10
5
25
13
12
Solution:
P(E)=
๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘กโ„Ž๐‘’ ๐‘’๐‘ฃ๐‘’๐‘›๐‘ก ๐‘๐‘Ž๐‘› ๐‘œ๐‘๐‘๐‘ข๐‘Ÿ
๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘œ๐‘ ๐‘ ๐‘–๐‘๐‘™๐‘’ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘ 
P(Shein)=
25
65
P(Shein)=
25
65
P(Shein)=
5
13
÷
5
5
2. Mario has 45 red chips, 12 blue chips, and 24 white chips. What is the probability that Mario
randomly selected a red chip?
Solution:
P(E)=
๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘กโ„Ž๐‘’ ๐‘’๐‘ฃ๐‘’๐‘›๐‘ก ๐‘๐‘Ž๐‘› ๐‘œ๐‘๐‘๐‘ข๐‘Ÿ
๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘œ๐‘ ๐‘ ๐‘–๐‘๐‘™๐‘’ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘ 
45
P(Red)=
81
P(Red)=
45
81
P(Red)=
5
9
÷
9
9
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