These 2 questions have to do with permutations
answer to #1-26
answer to #2-3439
1. In how many ways can you form a 3 digit number, using only the digits of the number 21 150?
2. How many natural numbers less than 10 000 include at least one nine.
please explain to me how you get the anwers
To find the solutions to these permutation problems, we need to use basic principles of counting.
1. In how many ways can you form a 3-digit number, using only the digits of the number 21 150?
To solve this, we need to consider three positions: the hundredth place, the tenth place, and the unit place.
We have two choices for the hundredth place, 2 and 1. Once we choose the digit for the hundredth place, we have five remaining digits to choose from for the tenth place. Similarly, for the unit place, we have four remaining digits to choose from.
So, the total number of ways we can form a 3-digit number is:
2 choices for the hundredth place * 5 choices for the tenth place * 4 choices for the unit place = 2 * 5 * 4 = 40
Therefore, there are 40 different ways to form a 3-digit number using only the digits 2, 1, 1, 5, and 0.
2. How many natural numbers less than 10,000 include at least one nine?
To find the solution to this problem, we can use the principle of complement.
First, let's find the total number of 4-digit natural numbers less than 10,000. We have 10 choices for each of the four positions. So, the total number of 4-digit natural numbers is:
10 choices for the thousandth place * 10 choices for the hundredth place * 10 choices for the tenth place * 10 choices for the unit place = 10 * 10 * 10 * 10 = 10,000
Next, let's find the number of 4-digit natural numbers that do not have any nines. For the thousandth place, we have 9 choices since we can't use zero. For the hundredth, tenth, and unit places, we have 10 choices for each.
So, the total number of 4-digit natural numbers that do not include any nines is:
9 choices for the thousandth place * 10 choices for the hundredth place * 10 choices for the tenth place * 10 choices for the unit place = 9 * 10 * 10 * 10 = 9,000
Finally, to find the number of 4-digit natural numbers that include at least one nine, we subtract the number of numbers that do not include any nines from the total number of numbers:
Total number of 4-digit natural numbers - Number of 4-digit natural numbers without any nines = 10,000 - 9,000 = 1,000
Therefore, there are 1,000 natural numbers less than 10,000 that include at least one nine.