## Permutations and Combinations Questions - 4

A committee of 6 members is to be selected from among 8 male and 6 female. Determine the number of ways of selecting equal number of male and female members in the committee ?
• 1100
• 1120
• 1150
• 1180
• 1260
Explanation

In a committee of 6 members having equal number of male and female members.

Then the committee consists with 3 male and 3 female members.

3 males can be selected in 8C3 ways.

3 female can be selected in 6C3 ways

Required number of ways = 8C3 x 6C3 = 56 x 20 = 1120 ways.

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A Question paper has two parts, Part A and Part B, containing 8 questions each, if the student has to choose 6 questions from each paper in how many ways can he choose the question ?
• 361 ways
• 289 ways
• 376 ways
• 784 ways
• 512 ways
Explanation

The student can chose question From Part A:

8C6 = $\displaystyle \frac{8!}{(8-6) \ !6!} = \frac{8 \times 7 \times 6!}{2! \times 6!} = \frac{58}{2}$ = 28 ways.

From Part B:

8C6 ways = 28 ways.

Total = 28 x 28 = 784 ways.

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In how many ways a student can answer 6 questions out of 10 questions ? If Question no 1 compulsorily be answered.
• 130
• 126
• 64
• 81
• 132
Explanation

If Question number 1 is compulsory the question can be chosen in 1C1 ways = 1.

Remaining 5 questions out of 9 questions in = 9C5 ways.

Therefore the student can chose the questions in 1C1 x 9C5 = 1 x 126 = 126 ways.

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A man has 3 pair of shoe, 5 shirts, 6 trousers, what are the number of ways in which he can dress himself with a combination of all the three ?
• 66
• 75
• 90
• 270
• 180
Explanation

He can dress in 3C1 x 5C1 x 6C1 = 3 x 5 x 6 = 90 ways.

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In how many ways can 8 identical presents be distributed among 4 children so that each children gets at least on present ?
• 36
• 24
• 45
• 35
• 28
Explanation

Formula => n â€“ 1Cr â€“ 1 [or] n â€“ 1 (Cr â€“ 1).

8 â€“ 1C4 â€“ 1 = 7C3 = $\displaystyle \frac{7 \times 6 \times 5 \times 4!}{4! \ 3 \times 2 \times 1}$ = 35.

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In a meeting of 10 persons, each person shakes hand with every other person. Determine the number of handshakes exchanged ?
• 45
• 50
• 90
• 120
• 35
Explanation

Number of shake hands = $\displaystyle \,{}_{10}C_2\, \ = \frac{10!}{(10-2)! \ 2!} = \frac{10 \times 9 \times 8!}{8! \ 2!}$ = 9 x 5 = 45 shake hands.

[OR]

Number of shake hands = $\displaystyle \frac{n (n â€“ 1)}{2} = \frac{10 (10 â€“ 1)}{2} = \frac{90}{2}$ = 45 shake hands.

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Seven identical coins are arranged in a row, determine the number of ways in which 4 heads and 3 tails can appear ?
• 10
• 26
• 21
• 14
• 35
Explanation

Required number of ways = $\displaystyle \frac{7!}{4! \ 3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1}$ = 35 ways.

Workspace
How many 6 digit numbers can be formed using the digits 3, 2, 1, 3, 2, 1 ?
• 720
• 90
• 49
• 80
• 120
Explanation

Number of digits = 6, 3 occupy 2 times, 2 occupy 2 times, 1 occupy 1 time.

Required number of ways = $\displaystyle \frac{6!}{2! \ 2! \ 2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2 \times 2 \times 2}$ = 90 ways.

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Find number of diagonals that can be drawn by joining vertices of a Hexagon ?
• 16
• 12
• 18
• 24
• 9
Explanation

Hexagon has 6 vertices, no three of them are collinear.

Any two points can be joined to form a straight line.

Number of straight lines which can be formed by joining these 6 points:

=> 6C2 = $\displaystyle \frac{6 \times 5 \times 4!}{4! \times 2!}$ = 15 straight lines.

Out of 15, 6 are sides = the number of diagonals = 15 â€“ 6 = 9.

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Find number of triangles that can be formed by joining vertices of a Pentagon by straight lines ?
• 9
• 8
• 15
• 10
• 25
Explanation

A pentagon has 5 vertices, and no three of them are collinear.

Triangles are formed by joining any three of the five vertices of the pentagon.

Required number of triangles = 5C3 = $\displaystyle \frac{5!}{(5-3)! \ 3!} = \frac{5 \times 4 \times 3!}{2! \ 3!}$ = 5 x 2 = 10.

10 triangles can be formed.

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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: