Summary
Highlights
This problem asks for the number of ways to arrange all letters in 'MATH'. It can be solved using the fundamental counting principle (4 choices * 3 choices * 2 choices * 1 choice = 24 ways) or the permutation formula (4P4), which simplifies to 4 factorial (4!).
This problem asks for the number of ways to arrange just two letters from 'MATH'. Using the fundamental counting principle, it's 4 choices * 3 choices = 12 ways. Alternatively, the permutation formula 4P2 gives the same result (4! / (4-2)! = 12).
Given 8 runners, how many ways can gold, silver, and bronze medals be awarded? Since the order of finishing matters (gold, silver, bronze are distinct), this is a permutation. It's calculated as 8P3 (8! / (8-3)!) which equals 8 * 7 * 6 = 336 ways. The fundamental counting principle also applies: 8 choices for gold, 7 for silver, 6 for bronze.
This problem involves forming teams of 5 from 20 basketball players where each can play any position. The order of selecting players for a team does not matter, making this a combination problem. The calculation is 20C5 (20! / (20-5)!5!), which results in 15,504 different teams.
Six letters (A, B, C, T, U, V) are in a hat. What's the probability of drawing exactly three letters that spell 'CAT'? Probability is defined as desired outcomes divided by total possible outcomes. There's 1 way to get the letters C, A, T. The total number of ways to choose 3 letters from 6, where order doesn't matter, is 6C3, which is 20. So, the probability is 1/20.
At a party with 30 people, if everyone shakes hands with everyone else, how many total handshakes occur? Since shaking hands between person A and B is the same as B and A, order doesn't matter. This is a combination problem: 30C2, which calculates to 435 handshakes.
100 people are running for a 4-person committee. What is the probability that you and your three friends (a group of 4) win the four seats? There's 1 way for your specific group of 4 to be chosen (4C4). The total ways to choose 4 people from 100 is 100C4, which is 3,921,225. So, the probability is 1/3,921,225.
What is the probability of being dealt a five-card hand that's all diamonds from a standard deck? There are 13 diamonds, so the number of ways to choose 5 diamonds is 13C5 (1,287). The total number of ways to choose any 5 cards from 52 is 52C5 (2,598,960). The probability is 1287/2598960.
In a class of 30 students, two names are drawn to lead a discussion. What is the probability that you and your best friend are chosen? There's 1 way for you and your friend to be chosen (2C2). The total number of ways to choose 2 students from 30 is 30C2, which is 435. So, the probability is 1/435.
How many distinct ways can all letters in 'GEOMETRY' be arranged? There are 8 letters in total. However, there are two 'E's. To account for this repetition, we calculate 8! divided by 2! (for the repeated 'E's). This gives 20,160 distinct arrangements.
How many four-digit numbers less than 7,000 can be formed if the number must be odd? The first digit can be 1-6 (6 choices). The last digit must be odd (1, 3, 5, 7, 9 - 5 choices). The middle two digits can be 0-9 (10 choices each). Multiplying these gives 6 * 10 * 10 * 5 = 3,000 numbers.
In how many ways can a 10-question true/false exam be answered, assuming all questions are answered? For each question, there are 2 choices (True or False). Since there are 10 questions, the total number of ways is 2 multiplied by itself 10 times, or 2^10, which equals 1,024 ways.
In how many ways can five people stand in a circle? While for a row it's 5! = 120, for a circle, rotations are considered the same arrangement. Therefore, we divide by the number of people, resulting in (n-1)! ways or n!/n. For 5 people, it's (5-1)! = 4! = 24 ways.
A shipment of 10 items has 3 defective ones (7 good). In how many ways can you receive 4 items with exactly 2 defective ones? This means you need to choose 2 defective items from the 3 available (3C2) AND 2 good items from the 7 available (7C2). Multiplying these combinations (3C2 * 7C2) gives 3 * 21 = 63 ways.