**Exercise 6.1**

Permutations and combinations

**Question and Answers**

**Class 11 – Maths**

Permutations and combinations

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Permutations and Combinations |

Chapter No. | Chapter 6 |

Exercise | Exercise 6.1 |

Category | Class 11 Maths NCERT Solutions |

**Question 1 How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that**

** (i) repetition of the digits is allowed?**

**Answer**

Let the 3-digit number be XYZ, where Z is at the units place, Y at the tens place and X at the hundreds place. When repetition is allowed, the units place can be filled in by any of the all 5 digits. Similarly, tens and hundreds digits can be filled by any of the all 5 digits. Thus, the number of ways in which 3-digit numbers can be formed from the given digits is 5×5×5 = 125.

**(ii) repetition of the digits is not allowed?**

**Answer**

Let the 3-digit number be XYZ, where Z is at the units place, Y at the tens place and X at the hundreds place. When repetition is not allowed, if units place is filled at first, then it can be filled by any of the given 5 digits. Thus, the number of ways is 5. Similarly, the tens place can be filled by remaining 4 digits and the hundreds place can be filled by remaining three digits. Therefore, by multiplication principle, the number of ways = 5×4×3 = 60.

**Question 2 How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?**

**Answer** Let the 3-digit number be XYZ, where Z is at the units place, Y at the tens place and X at the hundreds place. The units place can be filled in by any of the all 6 digits. Similarly, tens and hundreds digits can be filled by any of the all 6 digits, as repetition is allowed. Therefore, The total number of possible 3-digit numbers = 6 × 6 × 3 = 108

**Question 3 How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?**

**Answer **As repetition is not allowed, the first place can be filled in 10 different ways by any of the given 10 letters and second place can be filled in 9 different ways. Similarly third place can be filled in 8 different ways and the fourth in 7 different ways.

Therefore, the total number of 4-letter codes=10 × 9 × 8 × 7 = 5040

**Question 4 How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?**

**Answer** Given that 5 digit telephone number starts with 67. Let the five-digit number be PQRST. Therefore, the number is 67RST. As repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0,1,2,3,4,5,8,9. Thus, units place can be filled in 8 different ways, tens place can be filled in 7 different ways and hundreds place can be filled in 6 different ways.

∴ The total five-digit numbers with given conditions = 8 × 7 × 6 = 336

**Question 5 A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?**

**Answer **The possible outcomes after a coin toss are head and tail i.e., the number of ways of showing different face = 2

∴ The total number of possible outcomes after 3 times = 2 × 2 × 2 = 8

**Question 6 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?**

**Answer **The signal requires 2 flags. Given 5 flags of different colours.

The upper empty place can be filled in 5 different ways and the lower empty place can be filled in 4 different ways.

The number of ways in which signal can be given = 5 × 4 = 20

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