3)How mnay subsets of 6 integers taken from the numbers 1,2,3...,20 are
there such that there are no consecutive integers in any subset (e.g. if 5 is in the subset then 4 and 6 cannot be in it)?
This is a fairly challenging problem, what have you tried so far?
There are 20 choose 6 subsets, so I doubt you could use trial and error here.
To solve this problem, we need to find a systematic approach rather than using trial and error. Let's break it down step by step.
Step 1: Counting all subsets
The total number of subsets possible when choosing 6 integers from a set of 20 is given by the formula "n choose k" or nCk, which can be calculated as n! / (k!(n-k)!). In this case, n = 20 and k = 6. So, the total number of subsets is 20C6 = 20! / (6!(20-6)!) = 38,760.
Step 2: Counting subsets with consecutive integers
Now, we need to count the number of subsets that have consecutive integers. Let's try to visualize this problem to make it easier to understand.
Consider the integers 1, 2, 3, ..., 20. We can think of them as slots, where each slot can either be selected or not selected. To avoid consecutive integers, we need to skip choosing some numbers.
For example, let's say we choose the number 5. Then we cannot choose the numbers 4 and 6, as they are consecutive. So, we can only choose one of the following sets: {4, 6}, {4}, {6}, or {} (empty set).
If we choose the number 1, then we cannot choose 2, but we can choose any other remaining numbers, including 3. Similarly, if we choose 20, then we cannot choose 19, but we can choose any other remaining numbers.
Based on this observation, we can count the number of subsets with consecutive integers by considering how many possible choices we have for each number.
Step 3: Counting choices for each number
Let's analyze the choices for each number from 1 to 20.
For the first number, we have 20 choices (any number from 1 to 20). However, if we choose the first number as 20, then we cannot choose the second number (19). So, in this case, we have 19 remaining choices for the second number.
For the second number, if it is not chosen as 20, then we have 18 remaining choices, but if it is chosen as 20, then we have 17 remaining choices for the third number.
Continuing this process, for the sixth number, if the fifth and fourth numbers are chosen as 20 and 18, respectively, we have only 13 choices since we cannot choose 17 or any other larger number.
By multiplying the number of choices for each number, we can determine the total number of subsets with consecutive integers.
Step 4: Calculating the answer
To calculate the answer, we multiply the number of choices for each number for all possible cases where the sixth chosen number can vary.
In this case, the answer would be:
[(20 * 19 * 18 * 17 * 16 * 15) + (20 * 19 * 18 * 17 * 16 * 14) + ... + (20 * 19 * 18 * 14 * 13 * 12)], which can be simplified further if needed.
Calculating this value will give us the number of subsets that do not contain consecutive integers.
Keep in mind that this is a lengthy process, so using a programming language or a spreadsheet might be helpful to do the calculations accurately and efficiently.
I hope this explanation helps you understand the approach to solve this problem.