StudentAidUnit 7: Permutations and Combinations — Exercise 7 1
11th Class Mathematics · Unit 7: Permutations and Combinations
1.(i) Evaluate 10! / 0!8! (ii) Evaluate 12! / (3! × (12-3)!) (iii) Evaluate 1440 / (3! × 4!) + 2400 / (5! × 2!) (iv) Evaluate (n+2)! / (n+1)!
Given
(i) Expression
10!0! × 8!
Working
Since 0! = 1
= 10!1 × 8!
= 10 × 9 × 8!8!
= 10 × 9
Result
Result
= 90
Given
(ii) Expression
12!3!(12-3)!
Working
Simplify
= 12!3! × 9!
= 12 × 11 × 10 × 9!3 × 2 × 1 × 9!
= 12 × 11 × 106
= 13206
Result
Result
= 220
Given
(iii) Factorials
2! = 2, quad 3! = 6, quad 4! = 24, quad 5! = 120
Working
Expression
= 14406 × 24 + 2400120 × 2
= 1440144 + 2400240
= 10 + 10
Result
Result
= 20
Given
(iv) Expression
(n+2)!(n+1)!
Working
Expand
= (n+2)(n+1)!(n+1)!
Result
Result
= n + 2
2.(i) Write n³ - n in factorial form (ii) Write n(n-1)(n-2)...(n-r+1) in factorial form
Given
(i) Expression
n3 - n
Working
Factor
= n(n2 - 1)
= n(n-1)(n+1)
= (n+1)(n)(n-1)
= (n+1)!(n-2)!
Result
Factorial form
= (n+1)!(n-2)!
Given
(ii) Expression
n(n-1)(n-2)ldots(n-r+1)
Working
Multiply by (n-r)! / (n-r)!
= n(n-1)(n-2)ldots(n-r+1)(n-r)!(n-r)!
= n!(n-r)!
Result
Factorial form
= n!(n-r)!
3.Find n, if (n+4)! = 3024 · n!
Given
Equation
(n+4)! = 3024 · n!
Expand (n+4)!
(n+4)(n+3)(n+2)(n+1) · n! = 3024 · n!
Divide by n!
(n+4)(n+3)(n+2)(n+1) = 3024
Factor 3024
= 9 × 8 × 7 × 6
Compare with (5+4)(5+3)(5+2)(5+1)
(5+4)(5+3)(5+2)(5+1) = 9 × 8 × 7 × 6
Result
Solution
n = 5
4.If 1/7! + 1/8! = x/9!, find x
Given
Equation
17! + 18! = x9!
Working
Convert 1/7! to common denominator with 8!
17! + 18! = 88! + 18!
= 98!
Express in terms of 9!
98! = 9 × 99! = x9!
Result
Solution
x = 81
5.Prove that (2n+1)! / n! = [1 · 3 · 5 ... (2n-1)(2n+1)] · 2^n
Given
L.H.S.
(2n+1)!n!
Expand (2n+1)!
= (2n+1)(2n)(2n-1)(2n-2)ldots 6 · 5 · 4 · 3 · 2 · 1n!
Separate even and odd factors
= [(2n+1)(2n-1)ldots 5 · 3 · 1][(2n)(2n-2)ldots 6 · 4 · 2]n!
Factor even terms
= [1 · 3 · 5 ldots (2n-1)(2n+1)][2n · 2(n-1) ldots 2 × 3 · 2 × 2 · 2 × 1]n!
Simplify
= [1 · 3 · 5 ldots (2n-1)(2n+1)][n!][2n]n!
Result
R.H.S.
= [1 · 3 · 5 ldots (2n-1)(2n+1)] · 2n
6.Express as a single fraction: (n+2)! / (r+2)! + (n+1)! / (r+1)!
Given
Expression
(n+2)!(r+2)! + (n+1)!(r+1)!
Working
Factor out common terms
= (n+2)(n+1)!(r+2)(r+1)! + (n+1)!(r+1)!
= left[n+2r+2 + 1right] (n+1)!(r+1)!
= left[n+2+r+2r+2right] (n+1)!(r+1)!
= (n+r+4)(n+1)!(r+2)(r+1)!
Result
Single fraction
= (n+r+4)(n+1)!(r+2)!
7.There are four distinct colored balls and four boxes of same colors as those of the balls. Determine the number of possible ways the balls, one each in a box, can be placed such that a ball does not go to a box of its own color.
Given
Problem setup
Colors: Red (R), Yellow (Y), Green (G), Brown (B)
Case 1: Red ball swaps
Red ball in non-red box; number of remaining choices for 2 balls = 1
Example Case 1 arrangement
begin{array}{cc} Balls & R Y G B Boxes & Y R B G end{array}
Case 2: Red ball does not swap
Red ball not in red box, yellow ball not in red box; number of choices for remaining 3 balls = 2
Example Case 2 arrangements
begin{array}{cc} Balls & R Y G B Boxes & B R B G end{array} and other arrangements
Required number of ways
= 3(1 + 2)
Result
Total derangements
= 9