Permutations & Combinations
Miscellaneous Exercise Solutions
Q1
1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Vowels
DAUGHTER has 3 Vowels: {A, U, E}. Select 2: $^3C_2 = 3$.
Consonants
5 Consonants: {D, G, H, T, R}. Select 3: $^5C_3 = \frac{5 \times 4}{2} = 10$.
Selection
Ways to select letters = $3 \times 10 = 30$.
Arrange
Each selection of 5 letters can be arranged in $5! = 120$ ways.
Total
$30 \times 120 = 3600$.
3,600 words
Q2
2. How many words can be formed using all letters of EQUATION so that vowels and consonants occur together?
Groups
Vowels {E, U, A, I, O} (5 letters). Consonants {Q, T, N} (3 letters).
Logic
Treat (Vowels) as one block and (Consonants) as another block.
Blocks
Arrangement of 2 blocks = $2! = 2$.
Internal
Vowels arrange in $5! = 120$. Consonants arrange in $3! = 6$.
Total
$2 \times 120 \times 6 = 1440$.
1,440 words
Q3
3. Committee of 7 from 9 boys and 4 girls. Ways if: (i) exactly 3 girls (ii) at least 3 girls (iii) at most 3 girls.
(i) Exactly 3 girls
Select
3 Girls ($^4C_3$) AND 4 Boys ($^9C_4$).
Calc
$4 \times \frac{9 \times 8 \times 7 \times 6}{24} = 4 \times 126 = 504$.
(ii) At least 3 girls
Case 1
3 Girls, 4 Boys: $^4C_3 \times ^9C_4 = 4 \times 126 = 504$.
Case 2
4 Girls, 3 Boys: $^4C_4 \times ^9C_3 = 1 \times 84 = 84$.
Total
$504 + 84 = 588$.
(iii) At most 3 girls
Logic
Total selections ($^{13}C_7$) – (Selections with 4 girls).
Total
$^{13}C_7 = 1716$.
4 Girls
From (ii), Case 2 = 84.
Result
$1716 – 84 = 1632$.
(i) 504 (ii) 588 (iii) 1632
Q4
4. Words from EXAMINATION in dictionary order. How many before the first word starting with E?
Letters
A, A, E, I, I, M, N, N, O, T, X (11 letters).
Start with A
Fix A at start. Remaining 10 letters: {A, E, I, I, M, N, N, O, T, X}.
Permute
Repeats: 2 I’s, 2 N’s. Formula: $\frac{10!}{2!2!}$.
Calc
$\frac{3628800}{4} = 907200$.
Conclusion
Words starting with A come before E.
907,200 words
Q5
5. How many 6-digit numbers from 0, 1, 3, 5, 7, 9 divisible by 10 (no repeat)?
Condition
Divisible by 10 $\Rightarrow$ Unit digit must be 0.
Unit Place
Fixed as 0 (1 way).
Remaining
5 places left for {1, 3, 5, 7, 9}.
Permute
$^5P_5 = 5!$.
Result
$120$.
120 numbers
Q6
6. English alphabet (5 vowels, 21 consonants). Words with 2 diff vowels and 2 diff consonants?
Select V
2 from 5: $^5C_2 = 10$.
Select C
2 from 21: $^{21}C_2 = \frac{21 \times 20}{2} = 210$.
Combo
$10 \times 210 = 2100$ sets of 4 letters.
Arrange
Each set can be arranged in $4! = 24$ ways.
Total
$2100 \times 24 = 50400$.
50,400 words
Q7
7. Select 8 questions from 12 (Part I: 5, Part II: 7). At least 3 from each part.
Case 1
3 from I, 5 from II: $^5C_3 \times ^7C_5 = 10 \times 21 = 210$.
Case 2
4 from I, 4 from II: $^5C_4 \times ^7C_4 = 5 \times 35 = 175$.
Case 3
5 from I, 3 from II: $^5C_5 \times ^7C_3 = 1 \times 35 = 35$.
Total
$210 + 175 + 35 = 420$.
420 ways
Q8
8. 5-card combinations from 52 cards with exactly one king.
Select K
1 King from 4: $^4C_1 = 4$.
Select Others
4 others from 48 non-kings: $^{48}C_4$.
Calc
$^{48}C_4 = 194580$.
Total
$4 \times 194580 = 778320$.
778,320 combinations
Q9
9. Seat 5 men and 4 women in a row so women occupy even places.
Positions
Total 9 seats: 1 2 3 4 5 6 7 8 9.
Even
Even places are 2, 4, 6, 8 (4 seats). Women arrange in $4! = 24$.
Odd
Odd places are 1, 3, 5, 7, 9 (5 seats). Men arrange in $5! = 120$.
Total
$24 \times 120 = 2880$.
2,880 arrangements
Q10
10. From 25 students, choose 10. 3 students decide to either ALL join or NONE join.
Case 1
All 3 join: Need to choose remaining 7 from 22. $^{22}C_7 = 170544$.
Case 2
None join: Need to choose all 10 from 22. $^{22}C_{10} = 646646$.
Total
$170544 + 646646 = 817190$.
817,190 ways
Q11
11. Arrangements of ASSASSINATION so that all S’s are together.
Letters
A:3, S:4, I:2, N:2, T:1, O:1 (Total 13).
Group S
Treat {SSSS} as 1 unit.
New Count
1(S-group) + 3A + 2I + 2N + 1T + 1O = 10 units.
Permute
$\frac{10!}{3! \times 2! \times 2!}$ (Dividing by repeats of A, I, N).
Calc
$\frac{3628800}{6 \times 2 \times 2} = \frac{3628800}{24} = 151200$.
151,200 ways