Permutation Problems

Problem 1. How many 4-letter words, with or without meaning can be formed out of the letters of the word, ‘SATURDAY’ if repetition of letters is not allowed?

Solution:

Word SATURDAY has 8 letters i.e. S, A, T, U, R, D, A, and Y

To form 4-letter words, we first have to select 4 letters from these 8 letters

The ways of selecting 4 letters from 8 letters disregarding the order is ⁸C₄ .

After selection, there are 4! arrangements.

Hence, total number of words formed are: ⁸C₄ × 4!

Note: Selecting r objects out of n objects and then arranging them is same as r-permutation of n objects.

Problem 2. Find the number of ways of selecting 6 balls from 4 red, 6 blue, and 5 white given that the selection must have 2 balls of each color?

Solution:

We need to select 2 balls each of color red, blue and white as per the given condition.

Number of ways of selecting 2 red balls is ⁴C₂

Number of ways of selecting 2 blue balls is ⁶C₂

Number of ways of selecting 2 white balls is ⁵C₂

Hence, the total ways of selection are ⁴C₂ × ⁶C₂ × ⁵C₂  = 900

Problem 3. A class has just 3 seats vacant. Three people P, A, and R arrive at the same time. In how many ways can P, A, and R be arranged on those 3 vacant seats?

Solution:

For the very first seat, we have 3 choices i.e. P, A and R.

Let us randomly select A for the first seat.

For the second seat, we have 2 choices i.e. P and R

Let us randomly select R for the second seat.

For the third seat, we have 1 choice i.e. P

To summarize, we did the following:

Placed a person on seat 1 and placed a person on seat 2 and placed a person on seat 3.

Usage of and comes from the fact that occupation of all 3 seats was mandatory.

In mathematics, and is related with multiplication, hence we can say that total choices = 3 × 2 × 1 = 3!

If we change the seating order to P on the first seat, A on the second seat, and R on the third, does that change the total number of choices?

No, it does not. This is because equal importance is given to all three P, A, and R.

Problem 4. Find the number of ways of arranging 5 people if 2 of them always sit together?

Solution:

Let us consider the 2 people as a unit and the remaining 3 person as 3 separate units, So we have total 4 units.

Number of ways of arranging these 4 units is 4!

(just the way we proved in previous problem)

Number of ways of arranging the 2 person amongst themselves is 2!

In conclusion, the number of ways of arranging the 4 units and 2 person amongst themselves is 4! × 2!

Problem 5. Find all the three-letter words beginning and ending with a vowel. Given that repetition of alphabets is not allowed.

Solution:

Total vowels in english = 7 ( a, e, i, o, u, y, w)

Total consonants in english = 26 – 7 = 19

• Choices for the first letter are 7
• Choices for the third letter are 6 (since 1 vowel was placed as first letter)
• Choices for the middle letter are 19 + (7 – 2) = 24 (19 consonants + the vowels which were not placed)

Hence, total permutations are 7 × 6 × 24 = 1008

Do observe that here we first satisfied the vowel condition for first and third letter and then placed the middle letter.

Problem 6. An ice-cream shop has 10 flavors of ice cream. Find the number of ways of preparing an ice cream cone with any 3 different flavors?

Solution:

Let us consider n = 10 (total number of flavors) and r = 3 (number of different flavors needed)

For first flavor we have 10 choices

For second flavor we have 10 – 1 choices

For third flavor we have 10 – 2 choices and this is same as (n – r + 1)

The numbers of arrangement would be: 10 × (10 – 1) × (10 – 3 + 1) = 720

From this we can generalize that, the number of ways of arranging r objects out of n different objects is:

n × (n – 1)  . . . (n – r + 1) = ⁿP_r

Problem 7. 10 Olympians are running a race. Find the different arrangements of 1^st, 2^nd, and 3^rd place possible?

Solution:

We have to find different arrangements of 10 taken 3 at time.

Here,

• n = 10
• r = 3

Different arrangement of for 1, 2, and 3 places are

¹⁰P₃ = 10! / (7!)

= 10 × 9 × 8 = 720

Problem 8. How many even numbers lying between 1000 and 2000 can be formed using the digits 1, 2, 4, 5, and 9?

Solution:

Since the number is supposed to be even, the digits at units place must either be 2 or 4 leaving us with 2 choices for the digit at units place.

The number is supposed to lie between 1000 and 2000, So the digits at thousand’s place must be 1, we thus have 

1 choice for the digit at thousands place.

Rest of the 2 digits can be any one from 1, 2, 4, 5 and 9 i.e. 5 choices each

So the total permutations are: 2 × 5 × 5 × 1 = 50

Problem 9. How many 4 digit numbers divisible by 5 can be formed using 0, 3, 5, 7, and 9 if repetition of digits is not allowed?

Solution:

For the number to be divisible by 5, the digit at units place must either be 0 or 5, hence we have 2 possibilities.

Case 1. Digit at units place is 0

There are 4 choices for 10³ place (all numbers except 0)

There are 3 choices for the 10² place (1 got used up at 10³ place)

There are 2 choices for the 10¹ place (1 got used up at 10³ place and 1 at 10² place)

Hence the possible arrangements with 0 at units place are

4 × 3 × 2 = 24

Case 2. Digit at units place is 5

There are 3 choices for 10³ place (all except 0 and 5)

There are 2 choices for 10² place (1 got used up at 10³ place)

There is 1 choice for 10¹ place (1 got used up at 10³ place and 1 at 10² place)

Hence the possible arrangements with 5 at units place are 3 × 2 × 1 = 6

Total Arrangements = Number of arrangements in case 1 + Number of arrangements in case 2

Total Arrangements = 24 + 6 = 30

Permutation in mathematics is the arrangement of the object in a definite order. Permutation is similar to the combination and the basic difference between permutation and combination is that in permutation the order in which the object is taken is important while the combination is the arrangement of the objects when the order of the objects is not important.

Permutation is represented by the letter, P. For example permuation of set A = {1, 2, 3} when taken two object at a time is, {1, 2}, {1, 3}, {2, 3}, {3, 2}, {3, 1}, {2, 1}. In this article, we will learn about, permutation, its formula, examples, representation, properties, and types of permutation in detail.