arrangements with identical objects.pptx

SOUTHERN LUZON STATE UNIVERSITY ⃒ LABORATORY SCHOOL
slsu_labschool@slsu.edu.ph (042) 540-7576 / 0949-873-5043
Arrangements with “identical” objects
3rd Quarter

The Formula
The number of arrangements (in a row) of 𝑛 objects when 𝑛1 objects are
identical, 𝑛2 objects are identical, and so on … is:
𝑛!
𝑛1!×𝑛2!×⋯×𝑛𝑘!
Where 𝑛 = 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘

Example 1
How many possible arrangements of the letters of the word ASSESS are there?
Solution:
The word ASSESS has 6 letters and thus, we are going to arrange 6 objects in a
row. So, 𝑛 = 6. In that word, we have 4 identical objects (the 4 S’s). Hence, the
number of possible arrangements is:
6!
4!
= 30

Example 2
How many possible arrangements of the letters of the word MISSISSIPPI are
there?
Solution:
The word MISSISSIPPI has 11 letters. So, 𝑛 = 11. Among the 11 objects, the
following are identical: 4 I’s, 4 S’s, & 2P’s. So, the number of possible
arrangements is:
11!
4! × 4! × 2!
= 34,650

Example 3
An automobile dealer has 3 Fords, 2 Buicks, and 4 Dodges to place in the front
row of his car lot. In how many different ways by make of car can he display the
automobiles?
Solution:
The dealer has 9 cars to arrange. Thus, 𝑛 = 9. Assuming that the cars of the
same type are identical (3 Fords, 2 Buicks, and 4 Dodges), we have number of
possible ways:
9!
3!×2!×4!
= 1260

Example 4
In how many ways can we arrange the letters of the word FACTORIZED such that
the vowels must appear in the arrangement alphabetically?
NOTE:
The vowels in the said word are 𝐴, 𝑂, 𝐼 & 𝐸. These letters (wherever they are)
must appear in the following order:
𝐴 … 𝐸 … 𝐼 … 𝑂
EXAMPLES:
• DFACETRIZO
• TAEFDZICOR

Solution
We know that we have 4 vowels in consideration. Since they must appear in a
specific order, that is,
𝐴 … 𝐸 … 𝐼 … 𝑂
We will disregard (or cancel) any other configurations of these vowels. In other
words, we will treat the vowels as if they are identical. So, we have 4 “identical”
objects. The word FACTORIZED has 10 letters. So, 𝑛 = 10. Thus, the number of
possible arrangements is:
10!
4!
= 151,200

Example 5
In how many ways can we arrange the letters of the word FACTORIZED such that
the vowels, and consonants must appear in the arrangement alphabetically?
Solution:
The vowels in the word are 𝐴, 𝑂, 𝐼 & 𝐸. So they must appear in the following
order:
𝐴 … 𝐸 … 𝐼 … 𝑂 …
The consonants in the words are 𝐹, 𝐶, 𝑇, 𝑅, 𝑍 & 𝐷. So they must appear in the
following order:
𝐶 … 𝐷 … 𝐹 … 𝑅 … 𝑇 … 𝑍 …

Solution
Since the 4 vowels must appear in a specific order, treat them as if they are
“identical”. The logic applies to the 6 consonants. Therefore, the number of
possible arrangements is:
10!
4! × 6!
= 210

Example 6
Six boys and five girls are to be arranged a line. How many possible
arrangements can be made for them if the boys must be put in the line
according to their ascending height? Assume that boys have different heights.
Solution:
Since we are to arrange 6 boys and 5 girls, we have 11 persons to be arranged in
a line. So, 𝑛 = 11. Now, since the boys (with different heights)must appear in a
specific order, we can treat the 6 boys as if they are “identical”. Thus, we have
the number of possible arrangements:
11!
6!
= 55,440

Challenge Problem
Take a look again at Example 6. In that problem, we assumed that heights of the
boys are different. Consider and answer the following modified problem.
Six boys and five girls are to be arranged a line. How many possible
arrangements can be made for them if the boys must be put in the line
according to their ascending height? In this problem, assume that two of the
boys have the same height.

arrangements with identical objects.pptx

More Related Content

Similar to arrangements with identical objects.pptx

arrangements with identical objects.pptx