How many different number can be formed out of the digit of 111223? How many of these are greater than 300000
a) answer 60
Number. of different numbers formed =
6!/3!2!
=
6×5×4/2
=60 ways.
b)10
5!/3!2! = 120/6x2 = 10
Plz answer the question with a full solution
To find the number of different numbers that can be formed out of the digits of 111223, we can use the concept of permutations.
Step 1: Count the frequency of each digit in the given set.
- There are 3 occurrences of 1
- There are 2 occurrences of 2
- There are 1 occurrence of 3
Step 2: Calculate the total number of unique permutations using these digits.
- In this case, the total number of unique permutations is given by the formula:
N! / (n1! x n2! x n3! ...)
where N is the total number of digits and n1, n2, n3, ... are the frequencies of each digit.
Let's calculate it:
N = 6 (total number of digits)
n1 = 3 (frequency of digit 1)
n2 = 2 (frequency of digit 2)
n3 = 1 (frequency of digit 3)
Number of unique permutations = 6! / (3! x 2! x 1!) = 6! / (3 x 2 x 1) = 6 x 5 x 4 = 120
So, there are 120 different numbers that can be formed out of the digits of 111223.
To find how many of these numbers are greater than 300000, we need to consider the placement of the digits.
Step 3: Analyze the placement of digits in the numbers.
Since we want the numbers to be greater than 300000, the first digit has to be either 3 or larger.
- If the first digit is 3, the remaining 5 digits can be any permutation of the remaining available digits (i.e., 1, 1, 2, 2, and 3).
Number of permutations for this case = 5! / (2! x 2!) = 5! / (2 x 2) = 60
- If the first digit is 4 or larger, all the remaining 5 digits can be any permutation of the digits (i.e., 1, 1, 2, 2, and 3).
Number of permutations for this case = 5! / (2! x 2!) = 5! / (2 x 2) = 60
So, the total number of different numbers greater than 300000 is 60 + 60 = 120.
Therefore, out of the 120 different numbers that can be formed out of the digits of 111223, all of them are greater than 300000.
To find the number of different numbers that can be formed out of the digits of 111223, we need to consider the total number of digits and any repetitions.
The given set of digits is 111223.
Step 1: Count the total number of digits in the set.
In this case, there are 6 digits: 1, 1, 1, 2, 2, 3.
Step 2: Determine the number of repetitions for each digit.
In this case, the digit 1 repeats three times, digit 2 repeats twice, and digit 3 repeats once.
Step 3: Calculate the number of different numbers by considering permutations.
We need to divide the number of permutations of all the digits by the number of permutations of the repeating digits (to account for repetition).
The number of different numbers = 6! / (3! * 2! * 1!)
= (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1 * 2 * 1 * 1)
= 20
Therefore, there are 20 different numbers that can be formed using the digits from 111223.
To find how many of these numbers are greater than 300000, we need to consider the positions of the digits and the restrictions imposed by the given condition.
Following are the possibilities for the first digit:
1) If the first digit is 1, no number formed can be greater than 300,000.
2) If the first digit is 2, then all the remaining digits can be any of the numbers from the set 1, 1, 1, 2, 3. So, the number of different numbers greater than 300,000 formed using the digits is the same as the number of different numbers that can be formed from 1, 1, 1, 2, 3.
From our previous calculation, we know that the number of different numbers that can be formed from 1, 1, 1, 2, 3 is 20.
Therefore, there are 20 different numbers that can be formed from the digits 111223, and all of them are greater than 300,000 when the first digit is 2.
Lastly, if the first digit is 3, then no numbers formed can be greater than 300,000 because the given set does not contain a digit larger than 3.
Hence, the total number of different numbers greater than 300,000 that can be formed using the digits from 111223 is 20.