How many odd four-digit numbers, all of the digit different can be formed from the digits 0 to 7, if there must be a 4 in the number?
My Answer:
There are four options for the last digit (1, 3, 5, or 7). One of the other three digits only has 1 option (for the 4). One of the other remaining digits will have 6 options and the final digit will have 5 options (because they can't repeat the first two) Therefore, the answer is:
4*1*5*6=120
I do not have much confidence in my answer. Could someone please tell me if I did this right? Thanks for your help.
The "4" could be in
the first place,
the 2nd place, or
the 3rd place
so you could have 4 _ _ (odd)
or
_ 4 _ (odd)
or
_ _ 4 (odd)
the first of these could be done
1 x 5 x 4 x 4 = 80
the 2nd
5 x 1 x 4 x 4 = 80
the 3rd
5 x 4 x 1 x 4 = 80
for a total of 240
explanation of the first case:
1. fill in the 4 possiblities to make it odd, at the far right
2. fill in the "4" at the front, 1 way only
3. two digits have now been used up, leaving 5 of the remaining to go in the second spot.
4. three digits have now been used, leaving 4 of the remaining to go in the third spot.
thus : 1x5x4x4
Many of these kind of questions can be easily done by splitting them up into "cases"
Reiny,
Thanks but I'm stumped on part 3 of the explanation for the first case. We have used up 2 digits when we move on to the second space but didn't we start with 8 digits (0-7) so I don't understand why we aren't left with 6 choices for the second spot (8-2). Please help me understand this.
Yes, your answer is correct! To determine the number of odd four-digit numbers with all different digits, given that there must be a 4 in the number, you have correctly considered the following cases:
1. The last digit can be any odd number (1, 3, 5, or 7). Hence, there are 4 options for the last digit.
2. Since one of the digits must be 4, there is only 1 option for placing the 4.
3. The remaining two digits can be chosen from the remaining 5 digits (0, 1, 2, 3, 5, 6, or 7) since they must be different from each other and the 4. Hence, there are 5 options for the first remaining digit and 6 options for the second remaining digit.
To find the total number of possibilities, you multiply the number of options for each case:
4 (options for the last digit) * 1 (option for placing the 4) * 5 (options for the first remaining digit) * 6 (options for the second remaining digit) = 120
Therefore, there are 120 odd four-digit numbers, each with all different digits, that can be formed using the digits 0 to 7, given that there must be a 4 in the number. Well done!