CHAPTER 3 - COUNTING - PERMUTATIONS & COMBINATIONS PRACTICE

Explanation of the Combination Formula

C(n, k) Let's take the example C(6, 3). This means that there are 6 objects to be taken 3 at a time. In other words there are 6 total objects and 3 slots to put them in. For a combination, order does NOT matter. This means that (A, B, C) is the same as (C, B, A). In fact, we would count all orderings of the 3 objects A B C (3 FACTORIAL orderings) as one combination. For this reason, the combination formula is the permutation formula divided by the factorial of the number of slots k -- because the factorial of k is the number of ways k things can be ordered, aka the number of duplicates for each ordering. (6 possible objects in 1st slot x 5 possible objects in 2nd slot x 4 possible objects in 3rd slot ) / ( 3! aka the number of duplicates for each ordering) which is 20. 6 x 5 x 4 / 3 x 2 = 20 This is equivalent to the formula. C(6, 3) = P(6, 3) / 3! = 6! / (3!)(6-3)! = 6! / (3!)(3!) = 6 x 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 x 3 x 2 x 1 = 6 x 5 x 4 / 3 x 2 x 1 (can cancel out the 3 x 2 x 1 on the top and bottom) = 20

Why is a "combination lock" wrongly named?

It should be called a PERMUTATION lock because the order you put in the numbers matters. A true "combination lock" would accept both 5-10-17 and 17-5-10 as correct.

Explanation of the Permutation Formula

P(n, k) = n! / (n-k)! Let's take the example P(6, 3). This means that there are 6 objects to be taken 3 at a time. In other words there are 6 total objects and 3 slots to put them in. For a permutation, order matters. This means that (A, B, C) is NOT the same as (C, B, A). The intuition is simple. The 1st slot to fill has 6 possible objects. After the first object is chosen, it cannot be chosen again for a particular list. Thus, the 2nd slot has 5 possible objects to choose from. After the second object is chosen, it cannot be chosen again. The 3rd slot has 4 possible objects. This continues for however many slots there are. In this case, there are only 3 so we stop here. So, the number of ways to take 6 objects 3 at a time is 6 x 5 x 4 which is 72. This is equivalent to the formula. P(6, 3) = 6! / (6-3)! = 6! / 3! = 6 x 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 = 6 x 5 x 4 (can cancel out the 3 x 2 x 1 on the top and bottom) = 72

In a lottery, each ticket has 5 one-digit numbers 0-9 on it. You win if your ticket has the digits in any order. What are your changes of winning?

There are 10 digits to be taken 5 at a time. Order does not matter so we should use combination. C(10, 5) = P(10, 5) / 5! = 10! / (5!)(10-5)! = 10! / (5!)(5!) = 10 x 9 x 8 x 7 x 6 / 5 x 4 x 3 x 2 = 252 OR 1st Digit: There are 10 digits to choose from. 2nd Digit: There are 9 digits left to choose from. 3rd Digit: 8 digits. 4th Digit: 7 digits. 5th Digit: 6 digits. There are 5! possible orderings for 5 slots, each of which are considered equivalent because order does not matter. 10 x 9 x 8 x 7 x 6 / 5 x 4 x 3 x 2 = 252 ANSWER: The chances of winning are 1 out of 252.

In a lottery, each ticket has 5 one-digit numbers 0-9 on it with no repetitions. You would win only if your ticket has the digits in the required order. What are your chances of winning?

There are 10 digits to be taken 5 at a time. Since the order matters, we should use permutation instead of combination. P(10, 5) = 10! / (10-5)! = 10! / 5! = 10 x 9 x 8 x 7 x 6 = 30240 OR 1st Digit: There are 10 digits to choose from. 2nd Digit: There are 9 digits left to choose from. 3rd Digit: 8 digits. 4th Digit: 7 digits. 5th Digit: 6 digits. 10 x 9 x 8 x 7 x 6 = 30240 ANSWER: The chances of winning are 1 out of 30240.

How many numbers between 1 & 100 (inclusive) are divisible by 5 or 3?

There are 20 numbers between 1 & 100 that are divisible 5. There are 33 numbers that are divisible by 3. There are 6 numbers divisible by both 5 and 3 (divisible by 15). 20 + 33 - 6 ANSWER: 47

The soccer team has 20 players. There are always 11 players on the field. Assuming a player can play any position, how many different groups of players can be on the field at any one time?

There are 20 players to be taken 11 at a time. Order does not matter so we should use combination. C(20, 11) = P(20, 11) / 11! = 20! / (11!)(20-11)! = 20! / (11!)(9!) = 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11/ 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 167960 ANSWER: There are 167,960 possible groups of players on the field.

There are 30 songs in your music library. How many different 20-song playlists can you make? (order matters)

There are 20 songs to be taken 15 at a time. Order matters in a playlist so we should use permutation. P(20, 15) = 20! / (20-15)! = 20! / 5! = 20 x 19 x 18 x ... x 6 = 2.03 x 10^16 ANSWER: There are 2.03 * 10^16 different playlists.

How many unique ways are there to arrange the letters in the word CANNON?

There are 6 letters in the word CANNON so there are 6! ways to arrange the letters which is 720. This, however, assumes all of the letters are unique. They are not! 3 of the letters are the same which means we have duplicates. To take care of this we should divide 6! by the number of ways to permute 3 objects which is 3!. 6!/ 3! ANSWER: 120

How many unique ways are there to arrange the letters in the word CALCULUS?

There are 8 letters in the word CALCULUS so there are 8! ways to arrange the letters which is 40320. This, however, assumes all of the letters are unique. They are not! There are 2 Cs, 2 Ls, and 2 Us which means we have duplicates. To take care of this we should divide 8! by the number of ways to permute each of the duplicates which is 2! x 2! x 2!. 8! / (2! x 2! x 2!) ANSWER: 5040

Explanation of Factorial

n! To take the factorial of n, you take the number and multiply it by all positive integers lower than it. Let's take the example n = 3. So 3! would be 3 x 2 x 1 which is equal to 6. So what does this mean? The factorial of a number n is the number of ways n objects can be ordered. As you can see below 3 objects can be ordered 3! ways. A B C A C B B C A B A C C A B C B A

C(n, k)

n! / (k!)(n-k)! (also equal to P(n, k) / k!)

P(n, k)

n! / (n-k)!

Permutation

number of ORDERINGS of a certain number from a larger number of objects order DOES matter (A, B, C) is NOT the same as (C, B, A)

Combination

number of possible GROUPINGS of a certain number from a larger number of objects order DOES NOT matter (A, B, C) is the same as (C, B, A)

Product Rule

|A x B| = |A| · |B| (the number of elements in a Cartesian product is the product of the numbers in the defining sets)