Stochastic_Jimmy
New member
- Joined
- Jan 10, 2013
- Messages
- 27
So I'm struggling a bit with the following question:
How many 5 digit numbers can be made from the numbers 1 to 9, where no number can appear more than 3 times and at least one number must appear exactly twice?
I was thinking that there are 3 different cases that satisfy these requirements:
For Case 1 my thought was there are \(\displaystyle 9 \times 8 \) ways to get the two different numbers and then \(\displaystyle \binom{5}{3,2} \) arrangements for a total of \(\displaystyle 9 \times 8 \times \binom{5}{3,2}. \)
For Case 2 there are \(\displaystyle 9 \times 8 \times 7 \) ways to select the three different numbers and then \(\displaystyle \binom{5}{2,2,1} \) arrangements for a total of \(\displaystyle 9 \times 8 \times 7 \times \binom{5}{2,2,1}. \)
And for Case 3 there are \(\displaystyle 9 \times 8 \times \times 7 \times 6 \) ways to select the 4 different numbers and \(\displaystyle \binom{5}{2,1,1,1}. \) arrangements for a total of \(\displaystyle 9 \times 8 \times 7 \times 6 \times \binom{5}{2,1,1,1}. \)
Then the answer would just be the sum of the totals from these 3 cases.
Is this right or am I making an error somewhere?
Thanks so much!
How many 5 digit numbers can be made from the numbers 1 to 9, where no number can appear more than 3 times and at least one number must appear exactly twice?
I was thinking that there are 3 different cases that satisfy these requirements:
- 3 of one number and 2 of a different number (e.g. 11133 or 13311, etc.)
- 2 of one number, 2 of another number, and 1 other (e.g. 22334)
- 2 of one number and then 3 different numbers (e.g. 22345)
For Case 1 my thought was there are \(\displaystyle 9 \times 8 \) ways to get the two different numbers and then \(\displaystyle \binom{5}{3,2} \) arrangements for a total of \(\displaystyle 9 \times 8 \times \binom{5}{3,2}. \)
For Case 2 there are \(\displaystyle 9 \times 8 \times 7 \) ways to select the three different numbers and then \(\displaystyle \binom{5}{2,2,1} \) arrangements for a total of \(\displaystyle 9 \times 8 \times 7 \times \binom{5}{2,2,1}. \)
And for Case 3 there are \(\displaystyle 9 \times 8 \times \times 7 \times 6 \) ways to select the 4 different numbers and \(\displaystyle \binom{5}{2,1,1,1}. \) arrangements for a total of \(\displaystyle 9 \times 8 \times 7 \times 6 \times \binom{5}{2,1,1,1}. \)
Then the answer would just be the sum of the totals from these 3 cases.
Is this right or am I making an error somewhere?
Thanks so much!