Finite Mathematics Quiz: Permutations And Combinations
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Permutations And CombinationsQuestion 1 of 13

How many 6-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (without repetition) such that the number is divisible by 5 and the digits are in non-decreasing order?

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1
5
10
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Finite Mathematics Quiz

Finite Mathematics Quiz: Permutations And Combinations

Practice Permutations And Combinations in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Permutations And Combinations, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

How many 6-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (without repetition) such that the number is divisible by 5 and the digits are in non-decreasing order?

  1. 0 (correct answer)
  2. 1
  3. 5
  4. 10
Explanation: For a number to be divisible by 5, it must end in 0 or 5. Since we can only use digits 1, 2, 3, 4, 5, 6, the number must end in 5. For the digits to be in non-decreasing order, we need: d1≤d2≤d3≤d4≤d5≤d6d_1 ≤ d_2 ≤ d_3 ≤ d_4 ≤ d_5 ≤ d_6. Since the last digit must be 5, we need d1≤d2≤d3≤d4≤d5≤5d_1 ≤ d_2 ≤ d_3 ≤ d_4 ≤ d_5 ≤ 5. However, we must use each of the 6 digits exactly once (without repetition). If d6=5d_6 = 5, then we need to arrange digits 1, 2, 3, 4, 6 in non-decreasing order for the first 5 positions. This would require 1≤2≤3≤4≤6≤51 ≤ 2 ≤ 3 ≤ 4 ≤ 6 ≤ 5. But 6≤56 ≤ 5 is false since 6>56 > 5. Therefore, it's impossible to arrange the digits 1, 2, 3, 4, 6 in non-decreasing order when followed by 5. Hence, no such 6-digit numbers exist.

Question 2

A bookshelf contains 5 distinct fantasy books, 4 distinct mystery books, and 3 distinct science fiction books. In how many ways can a person choose 4 books to read, if they must select at least one fantasy book and at least one mystery book?

  1. 105
  2. 270
  3. 390 (correct answer)
  4. 495
Explanation: We use the complement rule. The total number of books is 5+4+3=125+4+3=12. The total number of ways to choose 4 books from 12 is (124)=495\binom{12}{4} = 495. The forbidden selections are those with no fantasy books OR no mystery books. Let AA be the set of selections with no fantasy books and BB be the set of selections with no mystery books. We want to find ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A \cup B| = |A| + |B| - |A \cap B|. ∣A∣|A| (no fantasy): Choose 4 from the 7 non-fantasy books (4 mystery, 3 sci-fi). (74)=35\binom{7}{4} = 35. ∣B∣|B| (no mystery): Choose 4 from the 8 non-mystery books (5 fantasy, 3 sci-fi). (84)=70\binom{8}{4} = 70. ∣A∩B∣|A \cap B| (no fantasy AND no mystery): Choose 4 from the 3 sci-fi books. (34)=0\binom{3}{4} = 0. The number of forbidden selections is 35+70−0=10535 + 70 - 0 = 105. The number of valid selections is the total minus the forbidden: 495−105=390495 - 105 = 390.

Question 3

A robot must travel from point A(0,0) to point B(6,4) on a Cartesian grid. The robot can only move one unit up (U) or one unit right (R) at each step. How many distinct paths are there from A to B that must pass through point P(3,2)?

  1. 20
  2. 100 (correct answer)
  3. 120
  4. 210
Explanation: The problem is broken into two independent parts: the path from A to P, and the path from P to B.
  1. Path from A(0,0) to P(3,2): This requires 3 Right moves and 2 Up moves, for a total of 5 moves. The number of distinct paths is the number of ways to arrange these moves: (52)=5!2!3!=10\binom{5}{2} = \frac{5!}{2!3!} = 10.
  2. Path from P(3,2) to B(6,4): This requires 6−3=36-3=3 Right moves and 4−2=24-2=2 Up moves, for a total of 5 moves. The number of distinct paths is (52)=5!2!3!=10\binom{5}{2} = \frac{5!}{2!3!} = 10. By the multiplication principle, the total number of paths from A to B passing through P is the product of the number of paths for each part: 10×10=10010 \times 10 = 100.

Question 4

A company has 10 software engineers. A team of 4 is to be selected for a new project. Among the engineers, there are 3 senior engineers: Alex, Ben, and Carla. How many teams can be formed if the team must include at least one of these three senior engineers?

  1. 35
  2. 175 (correct answer)
  3. 210
  4. 252
Explanation: This problem is most efficiently solved using the complement rule. First, calculate the total number of 4-person teams that can be formed from 10 engineers, without any restrictions. This is (104)\binom{10}{4}. Total teams = (104)=10×9×8×74×3×2×1=10×3×7=210\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210. Next, calculate the number of teams that have none of the specified senior engineers. This means selecting a team of 4 from the other 10−3=710 - 3 = 7 engineers. Teams with no seniors = (74)=7×6×53×2×1=35\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35. The number of teams with at least one senior engineer is the total number of teams minus the number of teams with no senior engineers: 210−35=175210 - 35 = 175.

Question 5

A company's board of directors consists of 7 men and 5 women. A 4-person subcommittee for a special project is to be formed. How many different subcommittees can be formed if the subcommittee must contain at most 2 men?

  1. 210
  2. 280
  3. 285 (correct answer)
  4. 495
Explanation: The condition "at most 2 men" means the subcommittee can have 0, 1, or 2 men. We calculate the number of ways for each case and sum them. Case 1: 0 men and 4 women. This requires choosing 0 from 7 men and 4 from 5 women: (70)(54)=1×5=5\binom{7}{0} \binom{5}{4} = 1 \times 5 = 5. Case 2: 1 man and 3 women. This requires choosing 1 from 7 men and 3 from 5 women: (71)(53)=7×10=70\binom{7}{1} \binom{5}{3} = 7 \times 10 = 70. Case 3: 2 men and 2 women. This requires choosing 2 from 7 men and 2 from 5 women: (72)(52)=21×10=210\binom{7}{2} \binom{5}{2} = 21 \times 10 = 210. The total number of possible subcommittees is the sum of these cases: 5+70+210=2855 + 70 + 210 = 285.

Question 6

A code consists of 5 symbols. The first 3 symbols must be uppercase letters from the English alphabet (A-Z), and the last 2 symbols must be digits (0-9). The code must satisfy the following conditions: all three letters are distinct, and the two digits are distinct. Additionally, exactly one of the letters must be a vowel (A, E, I, O, U). How many such codes are possible?

  1. 47,250
  2. 189,000
  3. 567,000 (correct answer)
  4. 1,404,000
Explanation: We calculate the number of possibilities for the letter part and the digit part separately. Letter Part (3 distinct letters, 1 vowel, 2 consonants):
  1. Choose which of the 3 positions will contain the vowel: (31)=3\binom{3}{1} = 3 ways.
  2. Choose which of the 5 vowels goes in that position: 5 ways.
  3. Fill the other two positions with distinct consonants (there are 21 consonants). The first can be chosen in 21 ways, the second in 20 ways. So, 21×20=42021 \times 20 = 420 ways. Total for letters: 3×5×21×20=6,3003 \times 5 \times 21 \times 20 = 6,300 ways. Digit Part (2 distinct digits):
  4. The first digit can be any of 10 digits: 10 ways.
  5. The second digit must be different from the first: 9 ways. Total for digits: 10×9=9010 \times 9 = 90 ways. Total Codes: Multiply the possibilities for both parts: 6,300×90=567,0006,300 \times 90 = 567,000.

Question 7

A bakery sells 6 different types of donuts. A customer wants to buy a dozen (12) donuts. How many different selections of 12 donuts are possible?

  1. 924
  2. 6,188 (correct answer)
  3. 12,376
  4. 18,564
Explanation: This is a problem of combinations with repetition, often solved using the "stars and bars" method. We are choosing k=12k=12 items (donuts) from n=6n=6 categories (types). The formula for combinations with repetition is (n+k−1k)\binom{n+k-1}{k} or (n+k−1n−1)\binom{n+k-1}{n-1}. Using the formula, we have n=6n=6 and k=12k=12. The number of possible selections is: (6+12−112)=(1712)=(175)\binom{6+12-1}{12} = \binom{17}{12} = \binom{17}{5} (175)=17×16×15×14×135×4×3×2×1=17×2×1×14×13=6,188\binom{17}{5} = \frac{17 \times 16 \times 15 \times 14 \times 13}{5 \times 4 \times 3 \times 2 \times 1} = 17 \times 2 \times 1 \times 14 \times 13 = 6,188

Question 8

How many distinct arrangements of the letters in the word ENGINEERING are possible if the three Es must be kept together?

  1. 15,120 (correct answer)
  2. 30,240
  3. 277,200
  4. 362,880
Explanation: The word ENGINEERING has 11 letters: E(3), N(3), G(2), I(2), R(1). To keep the three E's together, we can treat EEE as a single block. Now we are arranging 9 items: (EEE), N, N, N, G, G, I, I, R. The number of permutations of these 9 items, with repetitions of N (3 times), G (2 times), and I (2 times), is given by the multinomial formula: 9!3!2!2!=362,8806×2×2=362,88024=15,120\frac{9!}{3!2!2!} = \frac{362,880}{6 \times 2 \times 2} = \frac{362,880}{24} = 15,120

Question 9

A committee of 5 people is to be selected from a group of 6 married couples. What is the total number of committees that can be formed if no married couple is on the committee together?

  1. 192 (correct answer)
  2. 600
  3. 720
  4. 792
Explanation: To form a committee with no married couple, we must select 5 people from 5 different couples. First, choose which 5 of the 6 couples will contribute a member. This can be done in (65)\binom{6}{5} ways. Then, from each of these 5 chosen couples, we must choose one of the two people. This can be done in 252^5 ways. By the multiplication principle, the total number of committees is (65)×25=6×32=192\binom{6}{5} \times 2^5 = 6 \times 32 = 192.

Question 10

A manager needs to assign 5 different projects to 3 employees. Each employee must be assigned at least one project. In how many ways can the projects be assigned?

  1. 60
  2. 147
  3. 150 (correct answer)
  4. 243
Explanation: This is a surjective (onto) function problem. We use the Principle of Inclusion-Exclusion. First, the total number of ways to assign 5 distinct projects to 3 distinct employees without restrictions is 35=2433^5 = 243. From this, we subtract the assignments where at least one employee gets no projects. Let AiA_i be the set of assignments where employee ii gets no project. We want to find 35−∣A1∪A2∪A3∣3^5 - |A_1 \cup A_2 \cup A_3|. ∣A1∣|A_1|: All 5 projects go to the other 2 employees, so 25=322^5=32. Similarly, ∣A2∣=∣A3∣=32|A_2|=|A_3|=32. Sum is (31)×25=3×32=96\binom{3}{1} \times 2^5 = 3 \times 32 = 96. ∣A1∩A2∣|A_1 \cap A_2|: All 5 projects go to employee 3, so 15=11^5=1. Similarly, ∣A1∩A3∣=∣A2∩A3∣=1|A_1 \cap A_3|=|A_2 \cap A_3|=1. Sum is (32)×15=3×1=3\binom{3}{2} \times 1^5 = 3 \times 1 = 3. ∣A1∩A2∩A3∣|A_1 \cap A_2 \cap A_3|: All projects go to 0 employees, which is 0. By PIE, ∣A1∪A2∪A3∣=96−3+0=93|A_1 \cup A_2 \cup A_3| = 96 - 3 + 0 = 93. The number of valid assignments is 243−93=150243 - 93 = 150.

Question 11

In how many ways can 4 different math books and 3 different physics books be arranged on a shelf if no two physics books can be next to each other?

  1. 144
  2. 240
  3. 1,440 (correct answer)
  4. 5,040
Explanation: This is a spacing problem. First, arrange the 4 distinct math books. This can be done in 4!=244! = 24 ways. These books create 5 possible spaces where the physics books can be placed (one at each end and one between each pair of books): _ M _ M _ M _ M _. To ensure no two physics books are adjacent, we must place each of the 3 physics books in a different space. Since the physics books are distinct, the order in which we place them matters. We need to choose 3 of the 5 spaces and arrange the 3 books in them. This is a permutation: P(5,3)=5!(5−3)!=5×4×3=60P(5, 3) = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60. By the multiplication principle, the total number of arrangements is 4!×P(5,3)=24×60=1,4404! \times P(5, 3) = 24 \times 60 = 1,440.

Question 12

From a standard 52-card deck, a 5-card hand is dealt. How many distinct hands contain exactly one pair and three other cards of different ranks from each other and from the pair?

  1. 17,160
  2. 68,640
  3. 1,098,240 (correct answer)
  4. 1,349,088
Explanation: The construction of the hand can be broken into steps:
  1. Choose the rank for the pair (e.g., Kings). There are 13 ranks: (131)=13\binom{13}{1} = 13 ways.
  2. Choose 2 of the 4 suits for that rank: (42)=6\binom{4}{2} = 6 ways.
  3. Choose 3 other ranks for the remaining three cards. These must be different from the pair's rank and from each other. We choose 3 ranks from the remaining 12: (123)=220\binom{12}{3} = 220 ways.
  4. For each of these 3 ranks, choose 1 suit. There are 4 suits for each, so (41)×(41)×(41)=43=64\binom{4}{1} \times \binom{4}{1} \times \binom{4}{1} = 4^3 = 64 ways. By the multiplication principle, the total number of such hands is 13×6×220×64=1,098,24013 \times 6 \times 220 \times 64 = 1,098,240.

Question 13

A host is seating 8 guests around a circular table. Two of the guests, Alice and Bob, must not be seated next to each other. How many seating arrangements are possible?

  1. 1,440
  2. 3,600 (correct answer)
  3. 4,320
  4. 5,040
Explanation: This problem is best solved using the complement rule. First, find the total number of ways to seat 8 guests around a circular table, which is (8−1)!=7!=5,040(8-1)! = 7! = 5,040. Next, find the number of arrangements where Alice and Bob do sit together. Treat Alice and Bob as a single unit. Now we are arranging 7 entities (the Alice-Bob unit and the other 6 guests) in a circle, which can be done in (7−1)!=6!=720(7-1)! = 6! = 720 ways. The Alice-Bob unit can be arranged in 2!=22! = 2 ways (Alice on the left or Bob on the left). So, there are 720×2=1,440720 \times 2 = 1,440 arrangements where they are together. The number of arrangements where they are not together is the total minus this amount: 5,040−1,440=3,6005,040 - 1,440 = 3,600.