## Permutations and Combinations Questions - 1

In a class there are 15 boys and 20 girls, the teacher wants to select 1 boy and 1 girl to represent in a science fair. In how many ways can the teacher make this selection ?
• 35 ways
• 350 ways
• 300 ways
• 299 ways
• 275 ways
Explanation

Number of permutations when â€˜râ€™ objects are chooses out of â€˜nâ€™ different objects.

$\displaystyle \,{}_nP_r\,$ = $\displaystyle \frac{n!}{(n â€“ r)!}$

1 boy can be selected from 15 boys in 15P1 ways = $\displaystyle \frac{15!}{(15 â€“ 1)!} = \frac{15 \times 14!}{14!}$ = 15 ways.

1 girl can be selected from 20 girls in 20P1 ways = $\displaystyle \frac{20!}{(20 â€“ 1)!} = \frac{20 \times 19!}{19!}$ = 20 ways.

Total number of ways by fundamental principle of counting = m x n ways => mn ways => 15 x 20 = 300 ways.

Workspace
How many three digit numbers can be formed by using 1, 2, 3, 4 and 5 ?
• 275
• 5
• 25
• 125
• 150
Explanation

As there is no restriction on repetition, the numbers can be repeated.

There are 5 digits in 1, 2, 3, 4, 5.

We have to arrange 5 digits in 3 places [ here n = 5, r = 3 ]

Then, we can arrange units place digit in 5 ways, tens place digit in 5 ways, hundreds place digit in 5 ways.

Hence total number of required 3 digit number = 5 x 5 x 5 = 125 numbers.

[or] Directly => If repetition is allowed numbers that can be formed = nr => 53 = 125 numbers.

Workspace
How many numbers are there between 100 and 1000 such that 5 in the units place ?
• 120
• 90
• 195
• 145
• 125
Explanation

All these numbers have three digits.

5 is in the units place, the middle digit can be any one of the 10 digits, i.e 0 to 9.

Hundredths place digit can be any one of the 9 digits from 1 to 9.

Therefore hundreds place digit can be arranged in 9 ways, tens place digit can be arranged in 10 ways, units place digit can be arranged in 1 way [only with 5].

Required numbers = 9 x 10 x 1 = 90 ways.

Workspace
In how many ways 4 students sit in 5 vacant seats ?
• 20 ways
• 24 ways
• 60 ways
• 5 ways
• 120 ways
Explanation

$\displaystyle \,{}_nP_r\, = \,{}_5P_n\, = \frac{5!}{(5 â€“ 4)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{1!}$ = 120 ways.

Workspace
In how many ways the letters of the word "SIGNATURE" can be arranged, such that all the vowels be together ?
• 81 ways
• 729 ways
• 17280 ways
• 16000 ways
• 8100 ways
Explanation

The word "SIGNATURE" have 9 letters and 4 vowels and 5 consonants.

If all the vowels be together, then all vowels are considered as one unit.

S G N T R [A E I U]

5 consonants as 5 units total 5 + 1 = 6 units or objects. These can be arranged in 6! Ways.

And 4 vowels can be arranged among them self in 4! Ways.

Then total number of ways = 6! x 4! = 720 x 24 = 17280 ways.

Workspace
In how many ways letters of the word "SIGNATURE" can be arranged, such that on two vowels be together ?
• 15050 ways
• 43200 ways
• 15500 ways
• 17200 ways
• 22280 ways
Explanation

There are 9 letters in the word "SIGNATURE" with 5 consonants and 4 vowels.

If no two vowels be together, vowels can be placed in 6 places between consonants as shown below.

[Vowel = V, Consonant = C] => V C V C V C V C V C V.

The letters S G N T R can be arranged in 5! Ways, and 4 vowels in 6P4 ways

Total number of ways = 5! x 6P4 = 120 x 360 = 43200.   [6P4 = 360]

Workspace
How many four digit numbers greater than 4000 can be formed out of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 ?
• 4000
• 4500
• 4200
• 2016
• 2032
Explanation

Here the retraction is the number must be greater than 4000. Therefore the number in thousands place digit can be filled with 4, 5, 6, 7, 8 or 9. Then the first digit can be chosen in 6 different ways.

The rest of the digits can be chosen from the rest 8 numbers.

It can be in 8P3 ways = $\displaystyle \frac{8!}{( 8 â€“ 3 )!} = \frac{8 \times 7 \times 6 \times 5!}{5!}$ = 336.

Total number of 4 digit numbers greater than 4000 = 6 x 336 = 2016.

Workspace
The number of ways in which 8 boys sit around a table, so that two particular boys may sit together ?
• 1440 ways
• 120 ways
• 640 ways
• 240 ways
• 600 ways
Explanation

Circular permutation of â€˜nâ€™ objects = (n â€“ 1)!

Out of 8 boys two particular boys may sit together, there are (8 - 1) = 7 objects.

Then circular permutation = (7 â€“ 1)! = 6! Ways = 720 ways.

The two particular boys can be seated in 2! Ways.

Total number of ways 2! x 720 = 1440 ways.

Workspace
There are 8 trains plying between Town A to Town B. in how many ways can a person go from Town A to Town B, and return to Town A by a different train ?
• 64 ways
• 80 ways
• 48 ways
• 56 ways
• 72 ways
Explanation

He can go by 8 trains in 8 ways, and can return by 7 trains.

Then total number of ways = 8 x 7 = 56 ways.

Workspace
If 6 coins are tossed simultaneously, how many ways they can fall ?
• 128 ways
• 32 ways
• 64 ways
• 60 ways
• 48 ways
Explanation

As each coin has two faces, then total number of ways = 26 ways = 64 ways.

Workspace

#### Practice Test Report

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## Permutation and Combination Formulas

### Permutation:

Defination: The ways of arranging or selecting smaller or equal number of persons or objects from a group of persons or collection of objects with due regard being paid to the order of arrangement or selection is called Permutation.

### Important Rules:

1. Fundamental principle of counting: If an event can happen in exactly â€˜mâ€™ ways and if following it, a second event can happen in exactly â€˜nâ€™ ways then the two events in succession can happen in exactly â€˜m x nâ€™ => mn ways.

2. â€˜nâ€™ number of objects can be arranged in n! ways. n! read as â€˜n factorialâ€™.

n! denotes the product of the first â€˜nâ€™ natural numbers.

n! = n x (n â€“ 1) x (n â€“ 2) x (n â€“ 3) x ......

Ex: In how many ways 5 persons can be seated => Answer in 5! ways = 5 x 4 x 3 x 2 x 1 => 120 ways.

3. Number of permutations when â€˜râ€™ objects are chosen out of â€˜nâ€™ different objects.

$\displaystyle \,{}_nP_r\,$ = $\displaystyle \frac{n!}{(n â€“ r)!}$

Ex: How many 3 digit numbers can be formed out of the digits 1, 2, 3, 4, 5 and 6, if no digit is repeated in any number.

n = 6, r = 3, then $\displaystyle \,{}_6P_3\,$ = $\displaystyle \frac{6!}{(6 â€“ 3)!} = \frac{6 \times 5 \times 4 \times 3!}{3!}$ = 120 numbers.

4. Circular Permutation: When we have to arrange â€˜nâ€™ objects along a closed curve or along a circle. The number of circular permutations = ( n â€“ 1 ) ! ways

Ex: How many ways 5 persons can sit around a table?

Sol: ( 5 â€“ 1 )! = 4! = 4 x 3 x 2 x 1 = 24 ways.

5. Number of permutations when some of the things are alike, taken all at a time.

Note: 0! = 1

=> $\displaystyle \frac{n!}{n1!. n2!. n3! â€¦} \$.

### Combinations:

Defination: The number of ways in which smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not important.

1. Number of combinations = $\displaystyle \,{}_nC_r\,$ = $\displaystyle \frac{n!}{( n â€“ r )! r!}$

2. Combinations of n different things taking some or all of â€˜nâ€™ things at a time = ( 2n â€“ 1) ways.

Time: