Praxis Math Quiz: Apply Counting Principles
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Apply Counting PrinciplesQuestion 1 of 20

A student council consists of 6 seniors and 4 juniors. A 3-person subcommittee is to be formed. How many different subcommittees can be formed if the subcommittee must consist of either all seniors or all juniors?

120
80
24
144
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Praxis Math Quiz

Praxis Math Quiz: Apply Counting Principles

Practice Apply Counting Principles in Praxis Math 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 Apply Counting Principles, giving you a quick way to practice the rules, question types, and explanations that matter most for Praxis Math.

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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.

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Question 1

A student council consists of 6 seniors and 4 juniors. A 3-person subcommittee is to be formed. How many different subcommittees can be formed if the subcommittee must consist of either all seniors or all juniors?

  1. 120
  2. 80
  3. 24 (correct answer)
  4. 144
Explanation: This problem involves two mutually exclusive cases. The use of "either...or" means we should calculate the number of possibilities for each case and then add them together.\n- Case 1: The subcommittee consists of all seniors. We need to choose 3 seniors from 6. The number of ways is C(6,3)=6×5×43×2×1=20C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.\n- Case 2: The subcommittee consists of all juniors. We need to choose 3 juniors from 4. The number of ways is C(4,3)=4!3!1!=4C(4, 3) = \frac{4!}{3!1!} = 4.\n- Total number of possible subcommittees = (ways for Case 1) + (ways for Case 2) = 20+4=2420 + 4 = 24.

Question 2

There are 7 runners in a race. In how many different ways can the first, second, and third place medals be awarded if a specific runner, Alex, must finish in one of the top three positions?

  1. 210
  2. 30
  3. 180
  4. 90 (correct answer)
Explanation: We can solve this by considering the placement of Alex first.\n- Step 1: Choose a position for Alex. Alex can be 1st, 2nd, or 3rd, so there are 3 choices for Alex's position.\n- Step 2: Fill the other two medal positions. After placing Alex, there are 6 remaining runners and 2 remaining medal positions. The number of ways to arrange 2 runners from 6 is a permutation, P(6,2)=6×5=30P(6, 2) = 6 \times 5 = 30.\n- Step 3: Multiply the results. Total ways = (choices for Alex's position) ×\times (ways to fill other positions) = 3×30=903 \times 30 = 90.

Question 3

A license plate consists of 2 letters from the 26-letter alphabet followed by 3 digits from 0-9. How many unique license plates can be formed if the letters must be different from each other, but the digits are allowed to be repeated?

  1. 676,000
  2. 468,000
  3. 486,720
  4. 650,000 (correct answer)
Explanation: This problem uses the Fundamental Counting Principle with different conditions for letters and digits.\n- For the letters: The first letter has 26 choices. Since the letters must be different, the second letter has only 25 choices remaining. Ways to choose letters = 26×25=65026 \times 25 = 650.\n- For the digits: Each digit position has 10 choices (0-9) and repetition is allowed. Ways to choose digits = 10×10×10=1,00010 \times 10 \times 10 = 1,000.\n- Total unique license plates = 650×1,000=650,000650 \times 1,000 = 650,000.

Question 4

Using the digits {1, 2, 3, 4, 5} without repetition, how many 3-digit numbers can be formed that are greater than 300?

  1. 60
  2. 24
  3. 36 (correct answer)
  4. 48
Explanation: For a 3-digit number to be greater than 300, its first digit must be 3, 4, or 5. We can solve this by considering the choices for each position.\n- First digit (hundreds place): Must be 3, 4, or 5. There are 3 choices.\n- Second digit (tens place): After choosing the first digit, there are 4 digits remaining from the set {1, 2, 3, 4, 5}. So, there are 4 choices.\n- Third digit (units place): After choosing the first two digits, there are 3 digits remaining. So, there are 3 choices.\nBy the Fundamental Counting Principle, the total number of such numbers is 3×4×3=363 \times 4 \times 3 = 36.

Question 5

A school is forming a 6-member student advisory board from a pool of 8 seniors, 7 juniors, and 5 sophomores. If the board must have an equal number of members from each class, how many different boards can be formed?

  1. 5,880 (correct answer)
  2. 38,760
  3. 59
  4. 47,040
Explanation: A 6-member board with an equal number from each of the three classes means there must be 2 seniors, 2 juniors, and 2 sophomores. Since the order of selection does not matter, we use combinations. We calculate the number of ways to choose the members from each class and multiply the results.\n- Ways to choose 2 seniors from 8: C(8,2)=8×72=28C(8, 2) = \frac{8 \times 7}{2} = 28.\n- Ways to choose 2 juniors from 7: C(7,2)=7×62=21C(7, 2) = \frac{7 \times 6}{2} = 21.\n- Ways to choose 2 sophomores from 5: C(5,2)=5×42=10C(5, 2) = \frac{5 \times 4}{2} = 10.\n- Total number of different boards = 28×21×10=5,88028 \times 21 \times 10 = 5,880.

Question 6

A company has 10 employees. It needs to form two separate committees: a 3-person safety committee and a 2-person planning committee. If no employee can be on both committees, how many ways can the committees be formed?

  1. 2,520 (correct answer)
  2. 120
  3. 210
  4. 3,360
Explanation: This is a two-step selection process where order doesn't matter for the committees themselves, but the selections are dependent.\n- Step 1: Choose the 3-person safety committee from the 10 employees. This is a combination: C(10,3)=10×9×83×2×1=120C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ways.\n- Step 2: Since no employee can be on both committees, we must choose the 2-person planning committee from the remaining 10−3=710 - 3 = 7 employees. This is a combination: C(7,2)=7×62=21C(7, 2) = \frac{7 \times 6}{2} = 21 ways.\n- Step 3: By the Fundamental Counting Principle, multiply the results: 120×21=2,520120 \times 21 = 2,520.

Question 7

A password must be 5 characters long. The first two characters must be letters from the 26-letter alphabet, and the last three characters must be digits from 0 to 9. If letters can be repeated but digits cannot, how many different passwords can be created?

  1. 468,000
  2. 676,000
  3. 486,720 (correct answer)
  4. 1,370
Explanation: We use the Fundamental Counting Principle by considering the choices for each of the 5 positions.\n- 1st character (letter): 26 choices.\n- 2nd character (letter, repetition allowed): 26 choices.\n- 3rd character (digit): 10 choices (0-9).\n- 4th character (digit, no repetition): 9 choices remaining.\n- 5th character (digit, no repetition): 8 choices remaining.\nTotal number of passwords = 26×26×10×9×8=676×720=486,72026 \times 26 \times 10 \times 9 \times 8 = 676 \times 720 = 486,720.

Question 8

A pizza shop offers 8 different vegetable toppings and 5 different meat toppings. A customer wants to create a pizza with 3 vegetable toppings and 2 meat toppings. How many different pizzas can be made?

  1. 560 (correct answer)
  2. 1,287
  3. 6,720
  4. 66
Explanation: The order in which toppings are chosen does not matter, so this is a combination problem. We calculate the choices for vegetables and meat separately and then multiply them.\n- Step 1: Choose 3 vegetable toppings from 8. The number of ways is C(8,3)=8×7×63×2×1=56C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.\n- Step 2: Choose 2 meat toppings from 5. The number of ways is C(5,2)=5×42×1=10C(5, 2) = \frac{5 \times 4}{2 \times 1} = 10.\n- Step 3: The total number of different pizzas is the product of these two numbers: 56×10=56056 \times 10 = 560.

Question 9

A restaurant offers a three-course meal special. There are 4 appetizers, 6 main courses, and 3 desserts. A customer can choose to skip any one of the three courses, but must order at least two courses. How many different meal combinations are possible?

  1. 72
  2. 54
  3. 126 (correct answer)
  4. 114
Explanation: We need to consider the different cases for ordering at least two courses and sum the possibilities.\n- Case 1: Appetizer and Main only. Number of combinations = 4×6=244 \times 6 = 24.\n- Case 2: Appetizer and Dessert only. Number of combinations = 4×3=124 \times 3 = 12.\n- Case 3: Main and Dessert only. Number of combinations = 6×3=186 \times 3 = 18.\n- Case 4: All three courses (Appetizer, Main, and Dessert). Number of combinations = 4×6×3=724 \times 6 \times 3 = 72.\nTotal number of possible meals = 24+12+18+72=12624 + 12 + 18 + 72 = 126.

Question 10

A team of 4 is to be chosen from 9 employees. If two of the employees, Sarah and Tom, are feuding and cannot be on the team together, how many different teams can be formed?

  1. 126
  2. 105 (correct answer)
  3. 35
  4. 125
Explanation: This problem can be solved with complementary counting. First, find the total number of possible teams, then subtract the number of teams where Sarah and Tom are together.\n- Total teams of 4 from 9 employees (no restrictions): C(9,4)=9×8×7×64×3×2×1=126C(9, 4) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126.\n- Teams that include both Sarah and Tom: If we must include Sarah and Tom, we need to choose 2 more members from the remaining 7 employees. Ways = C(7,2)=7×62×1=21C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21.\n- Allowed teams = Total teams - Forbidden teams = 126−21=105126 - 21 = 105.

Question 11

How many 4-digit odd numbers can be formed using the digits 2, 3, 4, 5, and 6 if repetition of digits is not allowed?

  1. 120
  2. 60
  3. 24
  4. 48 (correct answer)
Explanation: This is a permutation problem with a restriction on the last digit. We should handle the restricted position first.\n- Step 1 (Units digit): For the number to be odd, the last digit must be odd. From the given digits {2, 3, 4, 5, 6}, the only odd digits are 3 and 5. So, there are 2 choices for the units digit.\n- Step 2 (Thousands digit): After choosing the units digit, there are 4 remaining digits. We can choose any of these for the first position. So, there are 4 choices.\n- Step 3 (Hundreds digit): There are now 3 digits remaining. So, there are 3 choices.\n- Step 4 (Tens digit): There are 2 digits remaining. So, there are 2 choices.\nTotal number of odd numbers = 4×3×2×2=484 \times 3 \times 2 \times 2 = 48.

Question 12

In how many ways can 4 math books and 3 science books be arranged on a shelf if all the science books must be kept together?

  1. 5,040
  2. 144
  3. 120
  4. 720 (correct answer)
Explanation: This is a permutation problem with a constraint. We can solve it in two steps.\n- Step 1: Treat the 3 science books as a single block. Now, we are arranging 4 math books and this 1 science block. The total number of items to arrange is 4+1=54 + 1 = 5. The number of ways to arrange these 5 items is 5!=1205! = 120.\n- Step 2: Within the science block, the 3 science books can be arranged among themselves. The number of ways to arrange these 3 books is 3!=63! = 6.\n- Step 3: To get the total number of arrangements, multiply the results from the two steps: 5!×3!=120×6=7205! \times 3! = 120 \times 6 = 720.

Question 13

A musician plans to perform 5 songs. If she has 7 songs to choose from, how many different sequences of 5 songs can she perform?

  1. 21
  2. 120
  3. 2,520 (correct answer)
  4. 16,807
Explanation: The question asks for the number of different sequences of songs. The word "sequence" implies that the order in which the songs are performed matters. Therefore, this is a permutation problem.\nWe need to find the number of permutations of choosing 5 songs from 7, which is denoted as P(7,5)P(7, 5).\nP(7,5)=7!(7−5)!=7!2!=7×6×5×4×3=2,520P(7, 5) = \frac{7!}{(7-5)!} = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2,520.

Question 14

A conference has 8 scheduled speakers. The keynote speaker must present first, and a special guest speaker must present last. In how many ways can the speaking order for the 8 speakers be arranged?

  1. 40,320
  2. 720 (correct answer)
  3. 5,040
  4. 120
Explanation: This is a permutation problem with fixed positions for two of the items.\n- There are 8 positions in the speaking order.\n- The first position is fixed for the keynote speaker (1 choice).\n- The last position is fixed for the special guest (1 choice).\n- This leaves 8−2=68 - 2 = 6 speakers to be arranged in the 66 remaining middle positions.\n- The number of ways to arrange these 6 speakers is 6!=6×5×4×3×2×1=7206! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.\n- The total number of arrangements is 1×6!×1=7201 \times 6! \times 1 = 720.

Question 15

From a group of 10 students, a president, vice president, and treasurer are to be chosen. In how many different ways can these positions be filled?

  1. 120
  2. 720 (correct answer)
  3. 1,000
  4. 30
Explanation: The problem asks for the number of ways to fill three distinct positions (president, vice president, treasurer). Because the positions are different, the order in which the students are chosen matters. Therefore, this is a permutation problem.\nWe are arranging 3 students from a group of 10. The number of permutations is calculated as P(10,3)=10!(10−3)!=10×9×8=720P(10, 3) = \frac{10!}{(10-3)!} = 10 \times 9 \times 8 = 720.

Question 16

A four-digit security code is created using the digits 0 through 9. If the first digit cannot be 0 and no digit can be repeated, how many different codes are possible?

  1. 5,040
  2. 4,536 (correct answer)
  3. 9,000
  4. 3,024
Explanation: This is a counting problem with restrictions. We have four positions to fill.\n- For the first digit, we can use any digit from 1 to 9 (since it cannot be 0). There are 9 choices.\n- For the second digit, we can use any digit from 0 to 9 except for the one used in the first position. Since we can now use 0, there are still 9 choices.\n- For the third digit, we have used two unique digits, so there are 8 choices remaining.\n- For the fourth digit, we have used three unique digits, so there are 7 choices remaining.\nBy the Fundamental Counting Principle, the total number of codes is the product of the choices for each position: 9×9×8×7=4,5369 \times 9 \times 8 \times 7 = 4,536.

Question 17

A committee of 4 members is to be selected from a group of 5 teachers and 8 students. If the committee must have exactly 2 teachers, how many different committees can be formed?

  1. 38
  2. 715
  3. 1,120
  4. 280 (correct answer)
Explanation: This problem involves two independent selections. Since the order of selection for a committee does not matter, we use combinations.\n- Step 1: Select 2 teachers from 5. The number of ways is C(5,2)=5!2!(5−2)!=5×42×1=10C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10.\n- Step 2: Select the remaining 2 members from the 8 students. The number of ways is C(8,2)=8!2!(8−2)!=8×72×1=28C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28.\n- Step 3: By the Fundamental Counting Principle, multiply the results of the two steps to find the total number of possible committees: 10×28=28010 \times 28 = 280.

Question 18

How many distinct arrangements can be made from the letters of the word "STATISTICS"?

  1. 3,628,800
  2. 50,400 (correct answer)
  3. 25,200
  4. 604,800
Explanation: This is a problem of finding permutations of a multiset (a set with repeated elements). The word "STATISTICS" has 10 letters in total.\n- The letter 'S' is repeated 3 times.\n- The letter 'T' is repeated 3 times.\n- The letter 'I' is repeated 2 times.\nThe formula for distinct arrangements is n!n1!n2!...nk!\frac{n!}{n_1! n_2! ... n_k!}, where n is the total number of letters and n1,n2,...n_1, n_2, ... are the frequencies of each repeated letter.\nSo, the number of distinct arrangements is 10!3!3!2!=3,628,800(6)(6)(2)=3,628,80072=50,400\frac{10!}{3!3!2!} = \frac{3,628,800}{(6)(6)(2)} = \frac{3,628,800}{72} = 50,400.

Question 19

Five friends (Amy, Ben, Carla, Dan, Eva) are sitting in a row of five seats. In how many arrangements will Amy and Ben not sit next to each other?

  1. 48
  2. 120
  3. 72 (correct answer)
  4. 96
Explanation: This problem is best solved using complementary counting: find the total number of arrangements and subtract the number of arrangements where Amy and Ben do sit together.\n- Total arrangements of 5 friends in 5 seats is 5!=1205! = 120.\n- To find arrangements where Amy and Ben sit together, treat them as a single block (AB). Now we are arranging 4 items: (the AB block, Carla, Dan, Eva). This can be done in 4!=244! = 24 ways.\n- Within the (AB) block, Amy and Ben can be arranged in 2!=22! = 2 ways (AB or BA).\n- Number of arrangements where they are together = 4!×2!=24×2=484! \times 2! = 24 \times 2 = 48.\n- Number of arrangements where they are NOT together = Total arrangements - Arrangements together = 120−48=72120 - 48 = 72.

Question 20

A book club has 8 fiction books and 6 non-fiction books. A member wants to borrow 3 books. How many selections are possible if the member wants to borrow either 3 fiction books or 3 non-fiction books?

  1. 364
  2. 1,120
  3. 76 (correct answer)
  4. 456
Explanation: This problem involves two mutually exclusive cases, so we calculate the possibilities for each case and add them. The order of selecting books does not matter, so we use combinations.\n- Case 1: Select 3 fiction books from 8. Ways = C(8,3)=8×7×63×2×1=56C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.\n- Case 2: Select 3 non-fiction books from 6. Ways = C(6,3)=6×5×43×2×1=20C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.\n- Total selections = (Ways for Case 1) + (Ways for Case 2) = 56+20=7656 + 20 = 76.