Evaluate (i) 8!, (ii) 4! - 3!.
Hint. Compute each factorial directly from its definition.
(i) 8! = 1x2x3x4x5x6x7x8 = 40320. (ii) 4!=24, 3!=6, so 4!-3!=24-6=18.
✦ Working through each part gives: (i) 40320. (ii) 18.
Is 3! + 4! = 7!?
Hint. Compute both sides independently before comparing -- factorials do not add the way their arguments do.
3!+4! = 6+24 = 30. 7! = 5040. Since 30 does not equal 5040, the statement is false.
✦ Working through each part gives: no, 3!+4!=30 but 7!=5040 -- they are not equal.
Compute 8!/(6! x 2!).
Hint. Cancel the shared 6! factor between 8! and the denominator before doing any large multiplication.
8!/6! = 8x7 = 56 (since 8!=8x7x6!). Dividing by 2!=2: 56/2=28.
✦ Working through each part gives: 28.
If 1/6! + 1/7! = x/8!, find x.
Hint. Rewrite both fractions on the left with a common denominator of 7! first, using 7!=7x6!.
1/6! + 1/7! = 7/7! + 1/7! = 8/7!. Setting this equal to x/8!: x = 8x8!/7! = 8x8 = 64 (since 8!=8x7!).
✦ Working through each part gives: x = 64.
Evaluate n!/(n-r)! when (i) n=6, r=2, (ii) n=9, r=5.
Hint. Cancel the shared (n-r)! factor to leave only the top few terms of n!.
(i) 6!/4! = 6x5 = 30. (ii) 9!/4! = 9x8x7x6x5 = 15120.
✦ Working through each part gives: (i) 30. (ii) 15120.
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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