Digits
How many odd four-digit numbers can we create from digits: 0, 3, 5, 6, 7?
(a) the figures may be repeated
(b) the digits may not be repeated
(a) the figures may be repeated
(b) the digits may not be repeated
Correct answer:
Showing 1 comment:
Martin
This is OK for the number of repeat combinations.
For the number of combinations without repetition, I have the following logic:
There are 3 options (3,5,7) in the 4th position (units), so we can choose from 3 digits.
In the 1st position (thousands) there are four options (3,5,6,7), but we used one digit in the 4th position, so we can choose from 3 digits.
In the 2nd position (hundreds) there are five options (0,3,5,6,7), but in the 1st and 4th positions we have already used two digits, so we can choose from 3 digits.
In the 3rd position (tens) there are five options (0,3,5,6,7), but in the 1st, 2nd and 4th positions we have already used three digits, so we can choose from 2 digits.
Calculation: 3x3x3x2 = 54.
Breaking down all the options would give the number of numbers for numbers starting with an odd number of 12 options, and for a number starting with 6 it is 18 options, ie 12x3 + 18 = 54. If we allowed the number to start at 0, it would be another 18 possibilities, which together would be 12x3 + 18x2 = 72, but if there were 0 instead of thousands, it would be three-digit numbers.
For the number of combinations without repetition, I have the following logic:
There are 3 options (3,5,7) in the 4th position (units), so we can choose from 3 digits.
In the 1st position (thousands) there are four options (3,5,6,7), but we used one digit in the 4th position, so we can choose from 3 digits.
In the 2nd position (hundreds) there are five options (0,3,5,6,7), but in the 1st and 4th positions we have already used two digits, so we can choose from 3 digits.
In the 3rd position (tens) there are five options (0,3,5,6,7), but in the 1st, 2nd and 4th positions we have already used three digits, so we can choose from 2 digits.
Calculation: 3x3x3x2 = 54.
Breaking down all the options would give the number of numbers for numbers starting with an odd number of 12 options, and for a number starting with 6 it is 18 options, ie 12x3 + 18 = 54. If we allowed the number to start at 0, it would be another 18 possibilities, which together would be 12x3 + 18x2 = 72, but if there were 0 instead of thousands, it would be three-digit numbers.
Tips to related online calculators
Would you like to compute count of combinations?
You need to know the following knowledge to solve this word math problem:
Related math problems and questions:
- Five-digit
Find all five-digit numbers that can be created from numbers 12345 so that the numbers are not repeated and then numbers with repeated digits. Give the calculation. - The largest number
Find the largest integer such that: 1. No figures is not repeat, 2. multiplication of every two digits is odd, 3. addition all digits is odd. - Unknown number 16
My number's tens is three times more then ones My number's ones is twice the number of thousands and my number's hundreds is half the number of of tens. I have two ones. which number am I? - Dd 2-digit numbers
Find all odd 2-digit natural numbers compiled from digits 1; 3; 4; 6; 8 if the digits are not repeated. - Three digits number
From the numbers 1, 2, 3, 4, 5 create three-digit numbers that digits not repeat and number is divisible by 2. How many numbers are there? - Three digits number 2
Find the number of all three-digit positive integers that can be put together from digits 1,2,3,4 and which are subject to the same time has the following conditions: on one positions is one of the numbers 1,3,4, on the place of hundreds 4 or 2. - Divisible by five
How many different three-digit numbers divisible by five can we create from the digits 2, 4, 5? The digits can be repeated in the created number. - How many
How many double-digit numbers greater than 30 we can create from digits 0, 1, 2, 3, 4, 5? Numbers cannot be repeated in a two-digit number. - Phone numbers
How many 7-digit telephone numbers can be compiled from the digits 0,1,2,..,8,9 that no digit is repeated? - Permutations
How many 4-digit numbers can be composed of numbers 1,2,3,4,5,6,7 if: and, the digits must not be repeated in the number b, the number should be divisible by 5 and the numbers must not be repeated c, digits can be repeated - Combinations
How many different combinations of two-digit number divisible by 4 arises from the digits 3, 5 and 7? - Digits
How many natural numbers greater than 4000 which are formed from the numbers 0,1,3,7,9 with the figures not repeated, B) How many will the number of natural numbers less than 4000 and the numbers can be repeated? - Two-digit number
In a two-digit number, the number of tens is three greater than the number of units. If we multiply the original number by a number written in the same digits but in reverse order, we get product 3 478. Find the original number. - Digits
How many five-digit numbers can be written from numbers 0.3,4, 5, 7 that is divided by 10, and if digits can be repeated? - Permutations with repetitions
How many times the input of 1.2.2.3.3.3.4 can be permutated into 4 digits, 3 digits and 2 digits without repetition? Ex: 4 digits = 1223, 2213, 3122, 2313, 4321. . etc 3 digits = 122.212.213.432. . etc 2 digits = 12, 21, 31, 23 I have tried permutation fo - Five-digit numbers
How many different five-digit numbers can be created from the numbers 2,3,5 if the number 2 appears in the number twice and the number 5 also twice? - Seven-segmet
Lenka is amused that he punched a calculator (seven-segment display) numbers and used only digits 2 to 9. Some numbers have the property that their image in the axial or central symmetry was again give some number. Determine the maximum number of three-di
