## Permutations and Combinations Questions - 3

Find the last digit of the sum of 1! + 2! + 3! + 4! +â€¦â€¦â€¦..100! ?
• 1
• 2
• 3
• 4
• 5
Explanation

All 5! and itâ€™s more have 0 as its units place digit.

Sum from 1! to 4! = 33.

Therefore units place digit will be 3.

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How many different numbers can be formed by using any four digits out of 1, 2, 3, 4, 5, and 6 ?. Begin with specific digit and no number being repeated in any number.
• 20
• 25
• 24
• 3
• 6
Explanation

One specific digit can be chosen in 1P1 ways, and remaining 3 digits in 5P3 ways.

Total => 1P1 x 5P3 = 1 x 20 = 20 ways.

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From a team of 16 players 12 players are to be chosen for a tourney in how many ways a selection board can make selection ?
• 1440
• 1640
• 720
• 5040
• 1820
Explanation

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

Required number of ways = $\displaystyle \,{}_{16}C_{12}\,$ = $\displaystyle \frac{16!}{(16 â€“ 12)! \ 12!}$

=> $\displaystyle \frac{164 \times 153 \times 147 \times 13 \times 12!}{4 \times 3 \times 2 \times 1 \times 12!}$ = 4 x 5 x 7 x 13 = 1820.

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A box contains 4 Red balls, 6 White balls, 5 Green balls, if 3 balls are drawn at random; determine the number of ways of selecting at least 1 Red ball in this selection ?
• 20 ways
• 220 ways
• 256 ways
• 324 ways
• 290 ways
Explanation

The box contains 4 Red balls and (6 + 5) = 11 non Red balls, if three balls are drawn Possibilities of selecting:

[i] At least 1 Red ball => 1 Red 2 non Red balls => number of ways = 4C1 x 11C2 = 4 x 55 = 220 ways.

[ii] 2 Red balls and 1 non Red ball = 4C2 x 11C1 = 6 x 11 = 66 ways.

[iii] 3 Red balls = 4C3 x 11C0 => $\displaystyle \frac{4!}{(4-3)! \ 3!}$ = 4 x 1 = 4 ways.

Total number of ways = 220 + 66 + 4 = 290.

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A box contain 3 Red balls 4 Blue balls, 6 White balls, if 3 balls are drawn, how many ways we can select 1 ball of each color ?
• 64 ways
• 72 ways
• 48 ways
• 144 ways
• 156 ways
Explanation

Red 3; Blue 4; White 6.

One is each color can be chosen in 3C1 x 4C1 x 6C1 = 3 x 4 x 6 = 72 ways.

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In how many ways a student can choose 4 subjects out of 8 subjects ?
• 78 ways
• 64 ways
• 72 ways
• 70 ways
• 128 ways
Explanation

The student can select 4 subjects out of 8 subjects.

In 8C4 ways = $\displaystyle \frac{8!}{(8 â€“ 4)! \ 4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!}$ = 2 x 7 x 5 = 70 ways.

Workspace
If two Dice are thrown simultaneously in how many ways can they be thrown ?
• 72 ways
• 6 ways
• 4 ways
• 2 ways
• 36 ways
Explanation

First dice is thrown in 6 ways; second dice is thrown in 6 ways.

By multiplication rule = m x n = mn ways

The dice can be thrown in 6 x 6 = 36 ways.

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A man wants to invite 6 friends, in how many ways can he invite one or more of his friends to a party ?
• 144
• 360
• 720
• 64
• 63
Explanation

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

He has to select one or more of his 6 friends in (26 â€“ 1) = 64 â€“ 1 = 63 ways.

Workspace
How many pairs of prime numbers are possible between 60 to 100 ?
• 18
• 28
• 33
• 32
• 36
Explanation

Prime numbers between 60 to 100 => 61, 67, 71, 73, 79, 83, 89, 97 => total 8 numbers.

Total number of pairs = 8C2 = $\displaystyle \frac{8!}{(8-2)! \ 2!} = \frac{8 \times 7 \times 6!}{6! \ 2!}$ = 28.

Workspace
Out of 5 consonants and 4 vowels, how many works can be made, each containing 3 consonants and 2 vowels ?
• 720
• 780
• 7200
• 7800
• 14400
Explanation

3 consonants can be selected out of 5 consonants in 5C3 ways.

2 vowels can be selected out of 4 vowels in 4C2 ways.

Hence required number of words, now each selection contains 3 consonants and 2 vowels total 5 letters, 5 letters can be arranged in 5! Ways.

Required number of ways = 5C3 x 4C2 x 5! = 10 x 6 x 120 = 7200 ways.

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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: