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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