Question
Please explain how it is done and difference of Permutations and Combination
t 6.7 (Counting Combinations). Let S= {1,2 100) How many 1. 10-combinations are there of S? 2. 10-combinations are there of S such that 2 does not appear? 10-combinations are there of S such that 2 does appear? 10-combinations are there of S that contain exactly four even numbers? 5. 10-combinations are there of S that contain at least eight even numbers? ombinations of S have no more than four elements total? Concept 6.6 (Analysis of if t provide the correct answer) olution Attempts). In each scenario, analyze your friend's explanation and decid is correct. If it is incorrect explain what their error is and how to solve the problem correctly (and 1. Problem: How many even integers are there in the set [1000, 1001, 1002,...,9999) with distinct (no repeated) digits? xplanation: "There are five ways to pick the ones digit since it can be either 0,2, 4,6, or 8 and the number must be even. Then there are eight ways to pick the thousands digit since it cannot be the digit we picked for the ones position and we can't lead with a zero. Then there are eight ways to pick the hundreds digit since it can be any digit except what was chosen for the ones and the thousands positions. Finally, there are seven ways to pick the tens digit since it can be any digit except what was chosen for the other three positions. Hence, by the Multiplication Principle, there are (5)(8)(8)(7-2, 240 integers of this form 2. Problem: How many subsets are there of the set {1,2-… 100) that have cardinality equal to 12 and have at least 9 even numbers? Explanation: "A subset of cardinality 12 is just a 12-combination since order does not matter. There are 50 even numbers in the set (1,2,... , 100) and the other half are odd. Hence there are () ways to select 9 even numbers from the set. Since we cannot have repeated numbers, there are 100-9-91 numbers left and since we need to pick 3 more to make a 12-combination, there are () ways to pick 3 more numbers. By t he Multiplication Principle, there are = 304,372,613,044,500 such subsets 3. Problem: How many 5-combinations of shirts and pants have at least three shirts if your closet contains 10 different shirts and 6 different pairs of pants? Explanation: "Any 5-combination with at least three shirts is the outcome of two events. Event one selects y ten total. There are ways to pick three shirts. Event two selects two more pieces f the remaining 13 pieces of clothing (7 shirts, 6 pants), and this can be done in (3) ways. Hence, by the Multiplication Principle, the answer is (3) (3) 9,360
Explanation / Answer
(According to Chegg policy, only one question with four subquestions will be answered. Please post the remaining in another question)
6.7: S = {1,2...100}
1. There are 100 numbers in all.
The number of 10-combinations of S are 100C10.
2. Since 2 does not appear, we are left with 99 numbers.
The number of 10-combinations of S such that 2 does not appear = 99C10.
3. Since 2 does appear, we have 99 other numbers out of which we need to choose 9 numbers as one number (2) is already chosen.
The number of 10-combinations of S such that 2 does appear = 99C9.
4. There are 50 even numbers and 50 odd numbers in S = {1,2.....100}.
The four even numbers can be chosen in 50C4 ways and the remaining 6 odd numbers can be chosen in 50C6 ways.
=> The number of 10-combinations of S such that there are exactly four even numbers = 50C6 * 50C4.
5. There are atleast 8 even numbers. Therefore we choose the 8 compulsory even numbers first.
This can be done in 50C8 ways.
Once these are selected, we will be left with 42 even and 50 odd numbers or 92 numbers in all.
We can choose the remaining two numbers in 92C2 ways.
=> The number of 10-combinations of S such that there are atleast 8 even numbers = 50C8 * 92C2.
6. The number of 0-combinations of S = 100C0.
The number of 1-combinations of S = 100C1.
The number of 2-combinations of S = 100C2.
The number of 3-combinations of S = 100C3.
The number of 4-combinations of S = 100C4.
=> The number of combinations of S which have no more than 4 elements = 100C0 + 100C1 + 100C2 + 100C3 + 100C4.
Remember, permutation counts order while combination does not. Permutation is arrangement and combination is mere selection.
1,2,3 is different from 1,3,2 in permutation but same in combination.