Explore basic concepts of permutations and combinations with this easy quiz, focusing on counting techniques, arrangements, and selections in everyday scenarios. Gain confidence in solving introductory problems related to combinatorics and probability.
If you have 4 different books, in how many ways can you arrange them on a shelf?
Explanation: There are 4 factorial or 4! ways to arrange 4 distinct books, which equals 24. Option 16 arises from incorrectly multiplying by 4, 3, 2 (ignoring the last book). Option 12 comes from miscounting the total arrangements. Option 20 does not match the calculation for any arrangement scenario with 4 books.
How many ways can you choose 2 students from a group of 5 for a project team?
Explanation: Choosing 2 out of 5 uses the combination formula C(5,2) which equals 10. Option 8 stems from subtracting possible pairs incorrectly. Option 12 could result from using the permutation formula (5*4=20) and dividing by 2 wrongfully. Option 15 is the result of misapplying permutation rules.
How many 3-digit numbers can be formed using the digits 1, 2, and 3 if digits can be repeated?
Explanation: With repetition allowed, each of the 3 places can be any of the 3 digits, so it's 3*3*3 = 27. Option 6 is the count if each digit is used only once. Option 9 results from considering 3 possibilities for the first digit and 3 for the next but stopping early. Option 12 does not fit the format for repeated choices.
In how many different ways can 3 medals (gold, silver, bronze) be awarded to 3 athletes?
Explanation: There are 3! or 6 ways to award medals to 3 athletes, as each medal must go to a different person, and order matters. Option 3 ignores the importance of order. Option 9 may confuse with some repeated assignments. Option 12 exceeds the possible arrangements for 3 items.
From a class of 7, how many ways can a committee of 4 students be formed?
Explanation: To form a group of 4 from 7 students, use combinations: C(7,4) = 35. Option 24 is from incorrectly multiplying 4 by 6. Option 84 is the number of ways if order mattered (permutations). Option 28 is the number from grouping 2 students instead of 4.
If you have 6 different crayons, how many ways can you pick 2 without caring about color order?
Explanation: Order does not matter, so use combinations: C(6,2) = 15. Option 12 treats it like arrangements (permutations) dividing incorrectly. Option 30 is the permutation answer (6*5) not divided by 2. Option 10 comes from subtracting instead of multiplying/combinating.
How many different ways can the letters in the word 'CAT' be arranged?
Explanation: The three unique letters can be arranged in 3! = 6 ways. Option 3 only counts single changes. Option 9 arises from overcounting possibilities by including repeated letters (which 'CAT' does not have). Option 5 is not a factorial and does not fit any arrangement count for three items.
In how many ways can 2 students from 5 stand in a line for a photo, if their order matters?
Explanation: Here, order is important, so use the permutation formula: 5 choices for the first spot and 4 for the second, giving 5*4 = 20. Option 10 is the combination count where order does not matter. Option 15 results from 5+4+3+2+1, which does not apply. Option 25 mistakenly multiplies 5 by itself.
From a deck of 10 uniquely numbered cards, how many ways can you select 3 cards if order does not matter?
Explanation: Order does not matter, so use the combination formula: C(10,3) = 120. Option 720 uses the permutation formula for ordered selections. Option 60 is an incorrect combination calculation. Option 210 may arise from using the formula for C(10,4) instead of C(10,3).
How many ways can six runners finish a race, assuming each gets a unique finishing position?
Explanation: All six runners get unique positions, so it’s 6! = 720 ways. Option 36 results from 6*6, which would be the case if runners could tie in positions. Option 64 is the result of 2^6, which does not apply here. Option 120 is 5! and would be correct for five runners, not six.