This is a question I made that I would like to solve.
Here are some points:
I am referring to combinations, not permutations. Therefore, whilst there are 2000 permutations of numbers between 1000 and 2999 inclusive, there should be less combinations. This is because the number 2324 has the same combination of numbers as 2342, just not in the same order, and should be considered the same combination of 4 digits. Order doesn't matter - it should solely depend on the different numbers obtained.
- As an example: The numbers 2341, 1432, 1342, 2431, 2143, and so on are all a combination of the four numbers, 1, 2, 3 and 4.
Each digit between 0 and 9 can repeat as many times as needed. (2222, 1111 is possible).
The 4-digit number must start with 1 or 2. This is really the only restriction.
Please alert me if there's something I need to clarify.
I've had a go, and tried researching, but I'm really lost...
I thought that dividing the total number of permutations, which is 2000, by possible repetitions would get the answer, such as by 2! for repetitions of 2 specific digits. But, this doesn't account for repetitions of 3 or 4 digits.
I just need a hint to help get there. I'm sure the solution could be smoother than I imagine.
$\endgroup$ 22 Answers
$\begingroup$For convenience of explanation, we change the range to $0000-1999$
With the first digit as $0$, we essentially want to count (say), weakly increasing three digit sequences, $1\leq x \leq y \leq z \leq9$
This is exactly equivalent to counting a strictly increasing sequence $1\leq x < (y+1) < (z+2) \leq 12= \binom{12}3 = 220$
and for the second sequence, we can neither use digit $0$ nor $1$
Thus ans $ =\binom{12}3 + \binom{11}3 = 385$
$\endgroup$ $\begingroup$First let’s find the number of combinations for $1000$ to $1999$. We have one $1$ and three arbitrary digits. The $1$ doesn’t make a difference; it juts gets added into any combination formed by the three other digits. So we want the number of combinations of three arbitrary digits. There are $10$ combinations with one distinct digit, $10\cdot9$ combinations with two distinct digits and $\binom{10}3=120$ combinations with three distinct digits, for a total of $220$.
We also have $220$ combinations for $2000$ to $2999$, for a total of $440$. Now we just have to substract the number of combinations we double-counted. These include a $1$ and a $2$ and two more digits. For the two other digits, there are $10$ combinations with one distinct digit and $\binom{10}2=45$ combinations with two distinct digits, for a total of $55$. (Here we have to use the unordered count $\binom{10}2$ because two distinct digits out of two digits are indistinguishable; above we had to use the ordered count $10\cdot9$ because two distinct digits out of three digits are distinguishable, with one occurring once and one occurring twice.)
So in total there are $220+220-55=385$ different combinations. The result is confirmed by this computer search.
$\endgroup$ 0
