# PE 3 - Permutation & Combination | Modern Math - Permutation & Combination

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

Out of 8 men and 6 women working in an office, 9 are to be chosen for a meeting including at least 3 men and 2 women. Two men, who were present in the last meeting, have to attend this meeting. Find the number of ways in which they can be selected.

- (a)
580

- (b)
680

- (c)
686

- (d)
786

Answer: Option D

**Explanation** :

Out of 8 men and 6 women, 9 people have to be selected such that 2 men must attend and there has to be at least 3 men and 2 women.

Since 2 men must attend, now we need to select at least 1 more man and 2 women out of remaining 6 men and 6 women.

**Case 1**: 1 man + 6 women

Number of ways = ^{6}C_{1} × ^{6}C_{6} = 6 × 1 = 6 ways

**Case 2**: 2 men + 5 women

Number of ways = ^{6}C_{2} × ^{6}C_{5} = 15 × 6 = 90 ways

**Case 3**: 3 men + 4 women

Number of ways = ^{6}C_{3} × ^{6}C_{4} = 20 × 15 = 300 ways

**Case 4**: 4 men + 3 women

Number of ways = ^{6}C_{4} × ^{6}C_{3} = 15 × 20 = 300 ways

**Case 5**: 5 men + 2 women

Number of ways = ^{6}C_{5} × ^{6}C_{2} = 6 × 15 = 90 ways

⇒ Total number of ways = 6 + 90 + 300 + 300 + 90 = 786.

Hence, option (d).

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

There are some cards. On each card 2-digit code is written selected from 0, 1, 2………9. But the card on which the code is written allows confusion between top and bottom, because these are indistinguishable. Thus, for example, the code 81 could be confused with 18. How many codes are there such that there is no possibility of any confusion?

- (a)
25

- (b)
75

- (c)
80

- (d)
None of these

Answer: Option C

**Explanation** :

Total number of 2-digit codes = 10 × 10 = 100

Digits which can be read upside down are 0, 1, 6, 8 and 9.

2-digit code formed using these digits that can be read upside down.

Number of 2-digit codes that can be formed using these 5 digits = 5 × 5 = 25.

Out these 25 codes there are some codes which will actually not create confusion: 00, 11, 88 , 69 and 96

Hence, (25 – 5 =) 20 codes can create confusion.

∴ Codes which will not create confusion = 100 – 20 = 80.

Hence, option (c).

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

Find the number of positive integral solutions for the equation x + y + z = 20 where x > 2, y > 3 and z > 4.

Answer: 45

**Explanation** :

Minimum value of x = 3, y = 4 and z = 5.

Let x = a + 3, y = b + 4 and z = c + 5 [a, b, c ≥ 0]

⇒ (a + 3) + (b + 4) + (c + 5) = 20

⇒ a + b + c = 8

∴ Number of integral solutions for x + y + z = 20 is same as number of integral solutions for a + b + c = 8.

⇒ Number of integral solutions for a + b + c = 8 = ^{8+3-1}C_{3-1} = ^{10}C_{2} = 45

Hence, 45.

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

Shubham has forgotten his friend’s 8-digit telephone number but remembers the following:

(a) The first 3 digits are either 305 or 936.

(b) The digit 0 occurs exactly three times and the digit 5 occurs exactly once.

(c) The number is an even number.

If Shubham wants to reach his friend, what is the maximum number of trials he has to make to be sure to succeed?

Answer: 3696

**Explanation** :

‘0’ occurs exactly 3 times and 5 occurs exactly once while the number is an even number.

**Case 1**: First 3 digits are 305

‘0’ has to occur 2 more times and one ‘5’ has already been used.

**Case 1(a)**: Last digit is ‘0’

Now, we have to use one more ‘0’ and other 3 digits can be any of 1, 2, 3, 4, 6, 7, 8, 9

3 0 5 __ __ __ _0_

One ‘0’ can be placed in 4C1 = 4 ways and remaining 3 digits can be filled in 8 × 8 × 8 = 512 ways.

∴ Total number of ways = 4 × 512 = 2048.

**Case 1(b)**: Last digit is not ‘0’

Now, we have to use two more ‘0s’ to be placed.

3 0 5 __ __ __ _2/4/6/8_

One ‘0’ can be placed in 4C2 = 6 ways,

Last digit can be filled with either 2 or 4 or 6 or 8 i.e., in 4 ways and

Remaining 2 digits can be filled in 8 × 8 = 64 ways.

∴ Total number of ways = 6 × 4 × 64 = 1536.

**Case 2**: First 3 digit are ‘936’

**Case 2(a)**: Last digit is ‘0’

Now, we have to use two more ‘0s’ and one ‘5’.

9 3 6 __ __ __ _0_

Two ‘0’ and one 5 can be placed in 4C3 × 3 = 12 ways and remaining 1 digit can be filled in 8 ways.

∴ Total number of ways = 12 × 8 = 96.

**Case 2(b)**: Last digit is not ‘0’

Now, we have to use three more ‘0s’ and one 5.

9 3 6 __ __ __ _2/4/6/8_

Last digit can be filled with either 2 or 4 or 6 or 8 i.e., in 4 ways and

Three ‘0s’ and a 5 can be placed in 4 ways,

∴ Total number of ways = 4 × 4 = 16.

⇒ Total number of combinations = 2048 + 1536 + 96 + 16 = 3696.

To make sure that Shubham reaches his friend, he will have to try a maximum of 3696 times.

Hence, 3696.

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

Everyday six visitors enter a residential complex and park their cars in one of the 10 adjacent parking spaces. The three lawyers among these five visitors arrive earlier than the other 3 visitors and never park in adjacent slots. Find the number of different ways in which the six cars can be arranged in the 10 spaces (linearly) such that there are no empty slots between any of the cars.

- (a)
720

- (b)
360

- (c)
480

- (d)
None of these

Answer: Option A

**Explanation** :

Let us first calculate the number of ways of arranging 6 cars linearly such that the three lawyers are never together.

3 cars of non-lawyers can be arranged in 3! = 6 ways.

| C | C | C |

Now, the cars of lawyers can be placed in between these 3 cars in ^{4}C_{3} × 3! = 24 ways

∴ Number of ways of arranging cars = 6 × 24 = 144 ways.

Parking slots ⇒ 1 2 3 4 5 6 7 8 9 10

Number of ways of selecting 6 adjacent parking slots = 5

∴ Total number of ways of parking these 6 cards in 6 adjacent slots = 5 × 144 = 720 ways.

Hence, option (a).

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

How many 4-digit numbers, that are divisible by 3, can be formed, using the digits 0, 1, 2, 3 and 4 if no digit is to occur more than once in each number?

Answer: 36

**Explanation** :

For a number to be divisible by 3, the sum of the digits should be divisible by 3.

∴ We need to select 4 numbers whose sum is divisible by 3.

We know 0 + 1 + 2 + 3 + 4 = 10,

Hence, excluding 1 or 4 will give the sum 9 or 6 (divisible by 3)

**Case 1**: 4 digits are 0, 2, 3 and 4

Number of 4-digit numbers = 3 × 3! = 18.

**Case 2**: 4 digits are 0, 1, 2 and 3

Number of 4-digit numbers = 3 × 3! = 18.

∴ Total number of numbers = 18 + 18 = 36.

Hence, 36.

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

If the letters of the word BELOW are arranged in a row in all possible ways and the arrangements are listed in alphabetical order as in a dictionary, then the rank of the word ELBOW is:

- (a)
32

- (b)
30

- (c)
31

- (d)
None of these

Answer: Option C

**Explanation** :

Letters when arranged in alphabetical order – B, E, L, O, W

To reach till the word ELBOW, we will consider all the words coming before it in alphabetical order.

Number of words starting with B = 4! = 24

Number of words starting with EB = 3! = 6

The next word will be E L B O W

∴ ELBOW will be the (24 + 6 + 1 =) 31st word.

Hence, option (c).

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

Find the sum of all possible 5-digit numbers that can be formed using the digits 1, 3, 5, 7 and 9 without repetition.

Answer: 66666600

**Explanation** :

Sum = 11111 × (1 + 3 + 5 +7 + 9) × 4! = 66666600.

Hence, 66666600.

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

In a regular polygon with 10 sides, find the number of triangles that can be formed with the vertices of the polygon, such that the triangles formed have at least one side in common with the polygon?

- (a)
120

- (b)
90

- (c)
60

- (d)
70

Answer: Option D

**Explanation** :

**Exactly one common side:**

We can chose one common side in 10 ways (any one of the 10 sides of the polygon)

If we chose two adjacent vertices of the polygon, we will get one side which will be common to polygon and the triangle. Now the next vertex should be chose such that we don’t get anymore common sides. Hence, the 3^{rd} vertex should not be adjacent to the ones already chose.

∴ The 3^{rd} vertex can be chose in 6 ways.

∴ Total number of such triangles = 6 × 10 = 60

**Exactly one common side:**

For two sides to be common we need to choose 3 adjacent vertices of the polygon. This can be done in 10 ways. (i.e., ABC, BCD, CDE, …, JAB)

∴ Total number of such triangles = 10

⇒ Total number of triangles = 60 + 10 = 70.

Hence, option (d).

Workspace:

**PE 3 - Permutation & Combination | Modern Math - Permutation & Combination**

There is an unlimited supply of identical red, blue and green coloured balls. In how many ways can 15 balls be selected from the supply?

Answer: 136

**Explanation** :

Let the number of red, blue and green balls selected be ‘r’, ‘b’ and ‘g’ respectively.

∴ r + b + g = 15

Since the balls of each color are identical, the question is same as distributing 15 identical objects in 3 distinct groups.

⇒ Number of ways = ^{15+3-1}C_{3-1} = 136.

Hence, 136.

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