How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Hint. This counts permutations of 9 distinct digits taken 3 at a time.
Using nPr with n=9, r=3: 9P3 = 9!/(9-3)! = 9x8x7 = 504.
✦ Working through each part gives: 504.
How many 4-digit numbers are there with no digit repeated?
Hint. The first digit cannot be zero, so handle it separately from the remaining three positions.
First digit: 9 choices (1-9, excluding 0). Second digit: 9 remaining choices (0-9 minus the first digit used). Third: 8 remaining. Fourth: 7 remaining. Total: 9x9x8x7=4536.
✦ Working through each part gives: 4536.
How many 3-digit even numbers can be made using the digits 1,2,3,4,6,7, if no digit is repeated?
Hint. Fill the units place first from the even digits available, then fill the remaining two positions from what's left.
Units digit must be even: 2, 4, or 6 (3 choices). Remaining 2 positions from the remaining 5 digits, no repetition: 5x4=20. Total: 3x20=60.
✦ Working through each part gives: 60.
Find the number of 4-digit numbers that can be formed using the digits 1,2,3,4,5 if no digit is repeated. How many of these will be even?
Hint. For the even count, fix the units digit first from the even digits available, then arrange the rest.
Total 4-digit numbers: 5P4=5x4x3x2=120. For even numbers, the units digit must be 2 or 4 (2 choices); the remaining 3 positions are filled from the remaining 4 digits: 4x3x2=24. Even total: 2x24=48.
✦ Working through each part gives: 120 total 4-digit numbers, of which 48 are even.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
Hint. This is an ordered selection of 2 people from 8, since the two roles are different.
Since chairman and vice chairman are distinct roles, order matters: 8P2 = 8!/(8-2)! = 8x7 = 56.
✦ Working through each part gives: 56.
Find n if (n-1)P3 : nP4 = 1 : 9.
Hint. Write both permutation terms in factorial form and simplify the ratio -- most of the factorial terms will cancel.
(n-1)P3 = (n-1)!/(n-4)!. nP4 = n!/(n-4)!. Ratio = (n-1)!/n! = 1/n. Setting 1/n = 1/9 gives n=9.
✦ Working through each part gives: n = 9.
Find r if (i) 5 x 4Pr = 6 x 5P(r-1), (ii) 5Pr = 6P(r-1).
Hint. Expand each side in factorial form, cancel the shared factorial term, and solve the resulting equation for r -- check any solution against the valid range of r afterward.
(i) 5x4!/(4-r)! = 6x5!/(6-r)!, simplifying to (6-r)(5-r)=6, i.e. r^2-11r+24=0, giving r=8 or r=3. Since 4Pr requires r<=4, r=8 is rejected, leaving r=3. (ii) 5!/(5-r)! = 6!/(7-r)!, simplifying to (7-r)(6-r)=6, i.e. r^2-13r+36=0, giving r=9 or r=4. Since 5Pr requires r<=5, r=9 is rejected, leaving r=4.
✦ Working through each part gives: (i) r = 3. (ii) r = 4.
Where students slip. Accepting both algebraic roots of the resulting quadratic without checking them against the permutation's valid domain (r must not exceed the smaller base number) leads to an incorrect extra answer.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Hint. Check whether any letters repeat before choosing a formula.
EQUATION has 8 distinct letters (E,Q,U,A,T,I,O,N), so the number of arrangements is simply 8! = 40320.
✦ Working through each part gives: 40320.
How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming no letter is repeated, if (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but the first letter is a vowel?
Hint. MONDAY has 6 distinct letters, 2 of them vowels (O, A). For part (iii), fix the first position with a vowel before arranging the rest.
(i) 6P4 = 6x5x4x3 = 360. (ii) 6P6 = 6! = 720. (iii) First letter: 2 choices (O or A). Remaining 5 positions from the remaining 5 letters: 5! = 120. Total: 2x120=240.
✦ Working through each part gives: (i) 360. (ii) 720. (iii) 240.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
Hint. Find the total number of distinct arrangements first, then subtract the arrangements where all four I's are glued together as one block.
MISSISSIPPI has 11 letters: I x4, S x4, P x2, M x1. Total arrangements: 11!/(4!4!2!) = 34650. Treating the 4 I's as one block leaves 8 objects (block + M,S,S,S,S,P,P) to arrange: 8!/(4!2!) = 840. Arrangements with I's not together: 34650-840=33810.
✦ Working through each part gives: 33810.
In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
Hint. PERMUTATIONS has 12 letters with T repeated twice. For (iii), think of P and S as occupying a pair of positions exactly 5 apart, then count how many such position-pairs exist along a row of 12.
(i) Fixing P and S at the two ends leaves 10 letters (including T twice) to arrange between them: 10!/2! = 1814400. (ii) The 5 vowels (E,U,A,I,O, all distinct) form one block; with the 7 consonants (including T twice) plus the block, that's 8 objects: 8!/2! x 5! = 20160x120 = 2419200. (iii) P and S must occupy positions i and i+5 for some i from 1 to 7 (7 position-pairs), and in either order (P first or S first), giving 14 placements; the remaining 10 letters (T twice) fill the other 10 positions in 10!/2! = 1814400 ways: 14x1814400=25401600.
✦ Working through each part gives: (i) 1814400. (ii) 2419200. (iii) 25401600.
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.
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