Apply Counting Principles - Praxis Core Math

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?
Answer: 24 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. - Case 1: The subcommittee consists of all seniors. We need to choose 3 seniors from 6. The number of ways is $C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ . - Case 2: The subcommittee consists of all juniors. We need to choose 3 juniors from 4. The number of ways is $C(4, 3) = \frac{4!}{3!1!} = 4$ . - Total number of possible subcommittees = (ways for Case 1) + (ways for Case 2) = $20 + 4 = 24$ .
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?
Answer: 486,720 We use the Fundamental Counting Principle by considering the choices for each of the 5 positions. - 1st character (letter): 26 choices. - 2nd character (letter, repetition allowed): 26 choices. - 3rd character (digit): 10 choices (0-9). - 4th character (digit, no repetition): 9 choices remaining. - 5th character (digit, no repetition): 8 choices remaining. Total number of passwords = $26 \times 26 \times 10 \times 9 \times 8 = 676 \times 720 = 486,720$ .
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?
Answer: 2,520 This is a two-step selection process where order doesn't matter for the committees themselves, but the selections are dependent. - Step 1: Choose the 3-person safety committee from the 10 employees. This is a combination: $C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ ways. - Step 2: Since no employee can be on both committees, we must choose the 2-person planning committee from the remaining $10 - 3 = 7$ employees. This is a combination: $C(7, 2) = \frac{7 \times 6}{2} = 21$ ways. - Step 3: By the Fundamental Counting Principle, multiply the results: $120 \times 21 = 2,520$ .
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?
Answer: 76 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. - Case 1: Select 3 fiction books from 8. Ways = $C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ . - Case 2: Select 3 non-fiction books from 6. Ways = $C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ . - Total selections = (Ways for Case 1) + (Ways for Case 2) = $56 + 20 = 76$ .
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?
Answer: 5,880 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. - Ways to choose 2 seniors from 8: $C(8, 2) = \frac{8 \times 7}{2} = 28$ . - Ways to choose 2 juniors from 7: $C(7, 2) = \frac{7 \times 6}{2} = 21$ . - Ways to choose 2 sophomores from 5: $C(5, 2) = \frac{5 \times 4}{2} = 10$ . - Total number of different boards = $28 \times 21 \times 10 = 5,880$ .
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?
Answer: 4,536 This is a counting problem with restrictions. We have four positions to fill. - For the first digit, we can use any digit from 1 to 9 (since it cannot be 0). There are 9 choices. - 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. - For the third digit, we have used two unique digits, so there are 8 choices remaining. - For the fourth digit, we have used three unique digits, so there are 7 choices remaining. By the Fundamental Counting Principle, the total number of codes is the product of the choices for each position: $9 \times 9 \times 8 \times 7 = 4,536$ .
How many distinct arrangements can be made from the letters of the word "STATISTICS"?
Answer: 50,400 This is a problem of finding permutations of a multiset (a set with repeated elements). The word "STATISTICS" has 10 letters in total. - The letter 'S' is repeated 3 times. - The letter 'T' is repeated 3 times. - The letter 'I' is repeated 2 times. The formula for distinct arrangements is $\frac{n!}{n_1! n_2! ... n_k!}$ , where n is the total number of letters and $n_1, n_2, ...$ are the frequencies of each repeated letter. So, the number of distinct arrangements is $\frac{10!}{3!3!2!} = \frac{3,628,800}{(6)(6)(2)} = \frac{3,628,800}{72} = 50,400$ .
A panel of 5 judges is to be selected from a group of 10 men and 7 women. How many different panels can be formed if the panel must contain at least one woman?
Answer: 5,936 This problem is best solved using complementary counting. First, find the total number of possible panels without any restrictions, then subtract the number of panels that violate the condition (i.e., panels with no women). - Total people = 10 men + 7 women = 17. - Total possible panels of 5 from 17 people is $C(17, 5) = \frac{17 \times 16 \times 15 \times 14 \times 13}{5 \times 4 \times 3 \times 2 \times 1} = 6,188$ . - The number of panels with no women is the number of ways to choose 5 judges from the 10 men: $C(10, 5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ . - The number of panels with at least one woman is Total - (panels with no women) = $6,188 - 252 = 5,936$ .
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?
Answer: 720 This is a permutation problem with fixed positions for two of the items. - There are 8 positions in the speaking order. - The first position is fixed for the keynote speaker (1 choice). - The last position is fixed for the special guest (1 choice). - This leaves $8 - 2 = 6$ speakers to be arranged in the $6$ remaining middle positions. - The number of ways to arrange these 6 speakers is $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ . - The total number of arrangements is $1 \times 6! \times 1 = 720$ .
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?
Answer: 280 This problem involves two independent selections. Since the order of selection for a committee does not matter, we use combinations. - Step 1: Select 2 teachers from 5. The number of ways is $C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$ . - Step 2: Select the remaining 2 members from the 8 students. The number of ways is $C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28$ . - Step 3: By the Fundamental Counting Principle, multiply the results of the two steps to find the total number of possible committees: $10 \times 28 = 280$ .
A bag contains 6 red marbles and 5 blue marbles. In how many ways can a person select 4 marbles such that at most 2 of them are red?
Answer: 215 The condition "at most 2 are red" means the selection can have 0, 1, or 2 red marbles. We calculate the number of ways for each case and add them up, ensuring each selection has a total of 4 marbles. - Case 1: 2 red marbles and 2 blue marbles. Ways = $C(6, 2) \times C(5, 2) = 15 \times 10 = 150$ . - Case 2: 1 red marble and 3 blue marbles. Ways = $C(6, 1) \times C(5, 3) = 6 \times 10 = 60$ . - Case 3: 0 red marbles and 4 blue marbles. Ways = $C(6, 0) \times C(5, 4) = 1 \times 5 = 5$ . Total ways = $150 + 60 + 5 = 215$ .
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?
Answer: 2,520 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. We need to find the number of permutations of choosing 5 songs from 7, which is denoted as $P(7, 5)$ . $P(7, 5) = \frac{7!}{(7-5)!} = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2,520$ .
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?
Answer: 720 This is a permutation problem with a constraint. We can solve it in two steps. - 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 $4 + 1 = 5$ . The number of ways to arrange these 5 items is $5! = 120$ . - 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 $3! = 6$ . - Step 3: To get the total number of arrangements, multiply the results from the two steps: $5! \times 3! = 120 \times 6 = 720$ .
Using the digits {1, 2, 3, 4, 5} without repetition, how many 3-digit numbers can be formed that are greater than 300?
Answer: 36 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. - First digit (hundreds place): Must be 3, 4, or 5. There are 3 choices. - 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. - Third digit (units place): After choosing the first two digits, there are 3 digits remaining. So, there are 3 choices. By the Fundamental Counting Principle, the total number of such numbers is $3 \times 4 \times 3 = 36$ .
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?
Answer: 720 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. We are arranging 3 students from a group of 10. The number of permutations is calculated as $P(10, 3) = \frac{10!}{(10-3)!} = 10 \times 9 \times 8 = 720$ .
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?
Answer: 105 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. - Total teams of 4 from 9 employees (no restrictions): $C(9, 4) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ . - 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 = $C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21$ . - Allowed teams = Total teams - Forbidden teams = $126 - 21 = 105$ .
A class has 8 boys and 6 girls. A group of 4 students is selected to represent the class. How many different groups can be formed if the group must contain at least 2 boys and at least 1 girl?
Answer: 756 The group of 4 must satisfy two conditions: at least 2 boys and at least 1 girl. We must consider all possible group compositions that meet these criteria and sum the possibilities. - Case 1: 2 boys and 2 girls. The number of ways to form this group is $C(8, 2) \times C(6, 2) = (\frac{8 \times 7}{2}) \times (\frac{6 \times 5}{2}) = 28 \times 15 = 420$ . - Case 2: 3 boys and 1 girl. The number of ways to form this group is $C(8, 3) \times C(6, 1) = (\frac{8 \times 7 \times 6}{3 \times 2 \times 1}) \times 6 = 56 \times 6 = 336$ . - A case of 4 boys and 0 girls would violate the 'at least 1 girl' condition. Total ways = (Ways for Case 1) + (Ways for Case 2) = $420 + 336 = 756$ .
Five committee members (A, B, C, D, E) are to be seated in five chairs arranged in a line. If member A must sit in the middle chair, how many seating arrangements are possible?
Answer: 24 This is a permutation problem with a restriction on one member's position. - The middle chair (the 3rd of 5 chairs) must be occupied by member A. There is only 1 way to place member A. - The remaining 4 members (B, C, D, E) can be arranged in the remaining 4 chairs. - The number of ways to arrange 4 people in 4 chairs is $4! = 4 \times 3 \times 2 \times 1 = 24$ . - The total number of arrangements is the product of the ways to place A and the ways to arrange the others: $1 \times 24 = 24$ .
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?
Answer: 560 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. - Step 1: Choose 3 vegetable toppings from 8. The number of ways is $C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ . - Step 2: Choose 2 meat toppings from 5. The number of ways is $C(5, 2) = \frac{5 \times 4}{2 \times 1} = 10$ . - Step 3: The total number of different pizzas is the product of these two numbers: $56 \times 10 = 560$ .
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?
Answer: 650,000 This problem uses the Fundamental Counting Principle with different conditions for letters and digits. - 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 = $26 \times 25 = 650$ . - For the digits: Each digit position has 10 choices (0-9) and repetition is allowed. Ways to choose digits = $10 \times 10 \times 10 = 1,000$ . - Total unique license plates = $650 \times 1,000 = 650,000$ .

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