Digits
How many odd four-digit numbers can we create from digits: 0, 3, 5, 6, and 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.
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