(a):how many digit number can be formed with the digit 2,2,2,3,3,4,5
(b):how many of these are greater than 3400000
(c):how many are greater than 3400000 and are divisible by 5
(d):how many are greater than 3400000 and are even
????
(a) 7!/(3!2!) = 420
(b) If 3,4 are used, that leaves 5!/3! = 20
(c) divisible by 5 means that 3,4,5 are used, leaving 4!/3! = 6
(2) divisible by 2 means that 2,3,4 are used, leaving 4!/2! = 12
Thanks steve
To answer these questions, we can use permutations and combinations. Let's break down each question and explain how to solve them step by step:
(a) How many digit numbers can be formed with the digits 2, 2, 2, 3, 3, 4, 5?
To find the number of unique combinations, we need to consider two scenarios:
1. If repetition of digits is allowed:
In this case, we can use the formula for combinations with repetition. The number of digit numbers that can be formed is given by:
n! / (n1! * n2! * n3! * ...),
where n is the total number of digits and n1, n2, n3, ... are the number of times each digit is repeated.
In this scenario, we have 7 digits (2, 2, 2, 3, 3, 4, 5), and the number 2 is repeated 3 times, the number 3 is repeated 2 times, and the remaining digits are unique.
So the number of digit numbers that can be formed is:
7! / (3! * 2! * 1! * 1! * 1!) = 7! / (3! * 2!) = (7 * 6 * 5) / (3 * 2) = 35
2. If repetition of digits is not allowed:
In this case, we can use the formula for permutations without repetition. The number of digit numbers that can be formed is given by:
n! / (n - r)!,
where n is the total number of digits and r is the number of digits needed for a number.
In this scenario, we have 7 digits (2, 2, 2, 3, 3, 4, 5), and we need a single-digit number.
So the number of single-digit numbers that can be formed is:
7! / (7 - 1)! = 7! / 6! = 7
(b) How many of these numbers are greater than 3400000?
To find the numbers greater than 3400000, we need to consider a few conditions:
1. The first digit must be 4.
2. The second digit must be greater than 3.
3. The remaining digits can be any of the given digits (2, 2, 2, 3, 3, 4, 5).
We have already calculated in part (a) that the total number of digit numbers that can be formed is 35.
To count the number of numbers greater than 3400000, we need to exclude the numbers that do not meet the conditions mentioned above.
The number of digit numbers greater than 3400000 is: 35 - 4! = 35 - 24 = 11
(c) How many of these numbers are greater than 3400000 and divisible by 5?
To find the numbers that are greater than 3400000 and divisible by 5, we need to consider an additional condition:
1. The last digit must be 5.
Using the same reasoning as in part (b), we calculated that the number of digit numbers greater than 3400000 is 11.
To count the number of numbers greater than 3400000 and divisible by 5, we need to exclude the numbers that do not meet the additional condition mentioned above.
The number of digit numbers greater than 3400000 and divisible by 5 is: 11 - 3! = 11 - 6 = 5
(d) How many of these numbers are greater than 3400000 and even?
To find the numbers that are greater than 3400000 and even, we need to consider an additional condition:
1. The last digit must be even, which means it can be 2 or 4.
Using the same reasoning as in part (b), we calculated that the number of digit numbers greater than 3400000 is 11.
To count the number of numbers greater than 3400000 and even, we need to exclude the numbers that do not meet the additional condition mentioned above.
The number of digit numbers greater than 3400000 and even is: 11 - 3! = 11 - 6 = 5
So, the answers to the given questions are:
(a) 35
(b) 11
(c) 5
(d) 5