Exercise 6.1 — Permutations and Combinations

1. 6.1.13 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   How many 3-digit numbers can be formed from the digits 1,2,3,4 and 5, assuming that (i) repetition of digits is allowed, (ii) repetition of digits is not allowed?

   Hint. Fill the three positions one at a time, counting how many digits remain available for each.

   (i) Each of the 3 positions independently has 5 choices, since repetition is allowed: 5x5x5=125. (ii) The first position has 5 choices, the second has 4 remaining, the third has 3 remaining: 5x4x3=60.

   ✦ Working through each part gives: (i) 125. (ii) 60.

2. 6.1.23 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

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

   Hint. Fill the units place first, since it has a restricted set of choices (only the even digits), then fill the remaining places freely.

   The units digit must be even: 2, 4, or 6 (3 choices). The other two positions can each be any of the 6 digits, since repetition is allowed: 6x6x3=108.

   ✦ Working through each part gives: 108.

3. 6.1.32 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

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

   Hint. Fill the 4 positions one at a time from the 10 available letters, reducing the choices by one each time.

   The first letter has 10 choices, the second has 9 remaining, the third has 8 remaining, and the fourth has 7 remaining: 10x9x8x7=5040.

   ✦ Working through each part gives: 5040.

4. 6.1.43 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   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?

   Hint. The first two digits are already fixed; only the remaining 3 positions need to be filled from the digits not yet used.

   With 6 and 7 fixed as the first two digits, 8 digits remain (0-9 excluding 6,7) for the remaining 3 positions, no repetition: 8x7x6=336.

   ✦ Working through each part gives: 336.

5. 6.1.52 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

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

   Hint. Each toss independently has 2 possible outcomes.

   By the multiplication principle: 2x2x2=8.

   ✦ Working through each part gives: 8.

6. 6.1.62 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   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?

   Hint. The upper flag position and the lower flag position are filled in succession from the available flags.

   The upper position has 5 choices; the lower position then has 4 remaining choices: 5x4=20.

   ✦ Working through each part gives: 20.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh106.pdf) — four numbered exercises (6.1-6.4, 31 questions) plus the chapter's Miscellaneous Exercise (11 questions); confirmed circular permutations do not appear anywhere in the chapter's theorems or its 42 questions, contrary to an earlier stub's claim about Exercise 6.3. Exercise 6.3 Q7's two-part nPr equation was solved with explicit domain checks (r must not exceed the smaller base number in each nPr term), rejecting an algebraically valid but combinatorially meaningless extraneous root in each part. Questions are referenced from the NCERT textbook for identification.