Permutations & Combinations
Exercise 6.3 • Permutations Formula
1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Given
Total digits $n = 9$ ({1, 2, …, 9})
Required
3-digit numbers ($r = 3$)
Formula
$^nP_r = \frac{n!}{(n-r)!}$
Apply
$^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!}$
Calculate
$9 \times 8 \times 7 = 504$
504 numbers
2. How many 4-digit numbers are there with no digit repeated?
Digits
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (Total 10)
Thousands
Can be 1-9 (9 options). Cannot be 0.
Remaining
9 digits left (including 0) for remaining 3 places.
Permute
$^9P_3 = 9 \times 8 \times 7 = 504$
Total
$9 \times 504 = 4536$
4536 numbers
3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Digits
{1, 2, 3, 4, 6, 7}. Total $n=6$.
Condition
Even Number $\Rightarrow$ Unit digit must be {2, 4, 6}. (3 choices)
Remaining
5 digits left for remaining 2 places.
Permute
$^5P_2 = 5 \times 4 = 20$
Total
$3 \times 20 = 60$
60 numbers
4. Find the number of 4-digit numbers using 1, 2, 3, 4, 5 (no repeat). How many are even?
Part 1
Total 4-digit numbers: $^5P_4 = 5 \times 4 \times 3 \times 2 = 120$
Part 2
Even numbers: Unit digit is {2, 4} (2 choices).
Rest
Remaining 4 digits for 3 places: $^4P_3 = 24$.
Even Total
$2 \times 24 = 48$
Total: 120, Even: 48
5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman?
Logic
Order matters (Chairman $\ne$ Vice Chairman) $\Rightarrow$ Permutation.
Calculation
$^8P_2 = \frac{8!}{(8-2)!} = \frac{8!}{6!}$
Expand
$8 \times 7 = 56$
56 ways
6. Find n if $^{n-1}P_3 : ^nP_4 = 1:9$.
Given
$\frac{^{n-1}P_3}{^nP_4} = \frac{1}{9}$
Expand
$\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$
Simplify
$\frac{(n-1)!}{n(n-1)!} = \frac{1}{9}$
Solve
$\frac{1}{n} = \frac{1}{9} \Rightarrow n = 9$
n = 9
7. Find r if (i) $^5P_r = 2 \cdot ^6P_{r-1}$
Expand
$\frac{5!}{(5-r)!} = 2 \times \frac{6!}{(7-r)!}$
Simplify
$\frac{1}{(5-r)!} = \frac{12}{(7-r)(6-r)(5-r)!}$
Quadratic
$(7-r)(6-r) = 12 \Rightarrow r^2 – 13r + 42 = 12$
Factor
$r^2 – 13r + 30 = 0 \Rightarrow (r-3)(r-10) = 0$
Check
$r$ must be $\le 5$, so $r=10$ rejected.
r = 3
8. How many words can be formed using all letters of EQUATION?
Letters
E, Q, U, A, T, I, O, N (8 letters, all unique).
Calculation
$^8P_8 = 8!$
Result
$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
40,320 words
9. Words from MONDAY. (i) 4 used (ii) All used (iii) First is vowel.
(i) 4 Used
$^6P_4 = 6 \times 5 \times 4 \times 3 = 360$
(ii) All
$^6P_6 = 6! = 720$
(iii) Vowel
Vowels: O, A (2 choices). Rest 5: $5!$.
Calc
$2 \times 120 = 240$
360; 720; 240
10. Permutations of MISSISSIPPI where four I’s do not come together.
Total
11! / (4! 4! 2!) = 34650
Together
Treat {IIII} as 1 unit. Total 8 units. 8! / (4! 2!) = 840.
Subtract
$34650 – 840$
33,810 words
11. PERMUTATIONS arrangements. (i) P start, S end (ii) Vowels together (iii) 4 letters between P and S.
(i) P…S
Fix P, S. Arrange 10 middle letters (2 T’s). $10!/2! = 1,814,400$.
(ii) Vowels
Vowels {E,U,A,I,O} = 1 unit. Total 8 units. $8!/2! \times 5! = 2,419,200$.
(iii) P..4..S
14 positions for P/S pair ($7 \times 2$). Middle 10 arranged. $14 \times 1814400 = 25,401,600$.
1,814,400; 2,419,200; 25,401,600