# PE 1 - Permutation & Combination | Modern Math - Permutation & Combination

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

In how many ways can 4 letters be selected from the letters of the word MISSISSIPPI?

- A.
25

- B.
20

- C.
19

- D.
21

- E.
None of these

Answer: Option D

**Explanation** :

There are 4 Ss, 4 Is, 2 Ps, 1 M

**Case 1**: All 4 distinct letters

There is only one way of selecting 4 distinct letters, i.e., S, I, P, M

**Case 2**: 2 similar letters and 2 other distinct letters

2 similar letters can be selected in 3 ways (i.e., S or I or P) and the remaining 2 distinct letters can be selected from other 3 letters in 3C2 = 3 ways.

∴ Total number of ways = 3 × 3 = 9 ways.

**Case 3**: 2 similar letters and 2 other similar letters

2 set of similar letters can be selected in 3C2 = 3 ways (i.e., SI or IP or PS)

**Case 4**: 3 similar letters and 1 other distinct letter

3 similar letters can be selected in 2 ways (i.e., S or I) and the remaining 1 distinct letter can selected from other 3 letters in 3 ways.

∴ Total number of ways = 2 × 3 = 6 ways.

**Case 5**: All 4 similar letters

4 similar letters can be selected in 2 ways (i.e., S or I)

⇒ Total number of ways = 1 + 9 + 3 + 6 + 2 = 21 ways.

Hence, option (d).

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

How many arrangements can be made by taking 4 letters from the word in the previous question?

Answer: 176

**Explanation** :

In the previous question we calculated number of ways of selecting 4 letters. Now for each of the cases discussed in the previous question we need to calculate the number of ways of arranging the letters.

**Case 1**: All 4 distinct letters

Number of ways of selection = 1

Number of ways of arrangementss = 1 × 4! = 24

**Case 2**: 2 similar letters and 2 other distinct letters

Number of ways of selection = 9

Number of ways of arrangements = 9 × $\frac{4!}{2!}$ = 108

**Case 3**: 2 similar letters and 2 other similar letters

Number of ways of selection = 3

Number of ways of arrangements = 3 × $\frac{4!}{2!2!}$ = 18

**Case 4**: 3 similar letters and 1 other distinct letter

Number of ways of selection = 6

Number of ways of arrangements = 6 × $\frac{4!}{3!}$ = 24

**Case 5**: All 4 similar

Number of ways of selection = 2

Number of ways of arrangements = 2 × $\frac{4!}{4!}$ = 2

⇒ Total number of arrangements = 24 + 108 + 18 + 24 + 2 = 176 ways.

Hence, 176.

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

A 5-ringed number lock has 10 digits on each ring. How many attempts at the maximum would have to be made before getting the right number?

- A.
10

^{5} - B.
5

^{10} - C.
10

^{5}- 1 - D.
5

^{10}- 1 - E.
8191

Answer: Option C

**Explanation** :

A number in each of the 5 rings can be selected in 10 different ways.

∴ Total number of combination possible = 10 × 10 × 10 × 10 × 10 = 10^{5}

Here, we don’t have to open the lock, we just have to figure out the right combination of lock.

If we consider that the last combination that we get is the right combination, then the maximum combinations we need to try would be 10^{5} – 1.

∴ Required number of attempts before getting the right combination = 10^{5} – 1

Hence, option (c).

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

In how many ways can 8 similar rings be worn on 5 fingers of a hand?

Answer: 126

**Explanation** :

We know, n similar objects can be distributed in r different groups in ^{n+r-1}C_{r-1} ways.

∴ 5 similar rings can be worn on 5 (different) fingers of a hand in ^{5+5-1}C_{5-1} = 126 ways.

Hence, 126.

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

The numbers at the extreme right and extreme left of a 6-digit telephone number with all different digits are 1 and 0 respectively. How many such telephone numbers are possible?

Answer: 1680

**Explanation** :

2 of the 6 digits of the number are already known.

Since all the digits in the number are different, the remaining 4 digits can be any digit except 1 and 0.

So, these 4 digits can be selected from any of the remaining 8 digits 2, 3, 4, 5, 6, 7, 8 and 9.

∴ Total number of possible telephone numbers = ^{8}P_{4} = 1680

Hence, 1680.

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

How many numbers greater than 24000 can be formed using the digits 7, 0, 9, 2 and 8 such that each digit is used exactly once?

- A.
48

- B.
72

- C.
45

- D.
90

- E.
96

Answer: Option D

**Explanation** :

The number must be greater than 24000 and each digit is to be used exactly once.

⇒ all the numbers will be 5 digits numbers.

So, numbers starting with 7, 8 and 9 will always be greater than 24000.

In this case the 1^{st} digit can be selected in 3 ways. The remaining 4 digits can be selected in 4! ways.

∴ Total number of possible numbers when the 1st digit is 7, 8 or 9 = 3 × 4 × 3 × 2 × 1 = 72.

Now, when the 1^{st} digit is 2, the second digit cannot be 0. The 2^{nd} digit can be any of 7, 8 or 9. So, the 2^{nd} digit can be selected in 3 ways. The remaining 3 digits can be selected in 3! ways.

So, total number of possible numbers when the 1^{st} digit is 2 = 1 × 3 × 3 × 2 × 1 = 18

The number cannot start with 0.

∴ Total number of required numbers = 72 + 18 = 90

Hence, option (d).

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

There are 10 pairs of socks of different colours. In how many ways can one choose socks of two different colours for both the legs?

Answer: 90

**Explanation** :

We will first select the right sock. This can be done in 10 ways.

The left sock can be any except that corresponding to the right leg already selected. So, it can be selected in 9 ways.

∴ Required number of ways = 10 × 9 = 90.

Hence, 90.

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

Among all convex polygons with n sides, the maximum number of points of intersection of the diagonals inside the polygon is 330. The value of n is

Answer: 11

**Explanation** :

By selecting 4 vertices we can get exactly one pair of diagonals which will intersect each other.

∴ The maximum number of points of intersection of the diagonals inside the polygon is same as selecting 4 vertices i.e., ^{n}C_{4}.

⇒ ^{n}C_{4} = 330

⇒ $\frac{n(n-1)(n-2)(n-3)}{4\times 3\times 2\times 1}$ = 330

⇒ n(n – 1) (n – 2)(n – 3) = 11 × 10 × 3 × 4 × 3 × 2

⇒ n(n – 1) (n – 2)(n – 3) = 11 × 10 × 9 × 8

⇒ n = 11

Hence, 11.

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

20 cities in a district are divided into 4 zones with five cities per zone. The telephone department of the district intends to connect the cities with telephone lines such that every two cities in the same zone are connected with two direct lines and every two cities belonging to different zones are connected with one direct line. How many direct telephone lines are required?

Answer: 230

**Explanation** :

**Connections within the Zone**

Number of connections required within 1 zone = ^{5}C_{2} × 2 = 20

∴ Number of connections required within each of 4 zones = 20 × 4 = 80 connections

**Connections across the Zones**

Number of connections required between any 2 zones = ^{5}C_{1} × ^{5}C_{1} = 25

Number of combination of zones = ^{4}C_{2} = 6

∴ Number of connections required across the 4 zones = 25 × 6 = 150 connections

⇒ Total number of connections required = 80 + 150 = 230.

Hence, 230.

Workspace:

**PE 1 - Permutation & Combination | Modern Math - Permutation & Combination**

There are certain number of students standing around a circle. Each possible pair of students not standing next to each other plays a kabaddi match in the middle of the circle for 2 minutes and 30 seconds. If the total time taken for all the matches is 135 minutes, then the number of students is equal to?

Answer: 12

**Explanation** :

Total number of pairs = ^{n}C_{2}

Number of pairs standing next to each other = n

∴ Number of pairs not standing next to each other = ^{n}C_{2} – n = $\frac{n(n-3)}{2}$

∴ Total time required = $\frac{n(n-3)}{2}\times \frac{5}{2}$ = 135

⇒ n(n – 3) = 108 = 12 × 9

⇒ n = 12

∴ There were 12 students standing in the circle.

Hence, 12.

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