My logic is since $3$ out of $4$ elements are chosen, each element would appear once. So a sequence would look like: $a\,b\,c\,x\,x\,x\,x\,x\,x\,x$
We have $7$ spots $x$ that can be whatever elements so: $3^7$
Choosing $3$ out of $4$ elements: $4\choose 3$$ = 4$ So in total, there are $4\cdot(3^7)$ sequences
Can someone tell me if this is right or not? Thanks
On
Use $\binom{4}{3}$ to make the selection of $3$ elements to show up in the sequence.
Now you have to ask, "How many of each should there be?" Since there needs to be at least one of each, we are looking for how many ways there are to partition $10$ into the sum of $3$ positive integers. There are probably fast ways to do that, but I just wrote them all down by hand:
$\{8,1,1\}$, $\{7,2,1\}$, $\{6,2,2\}$, $\{6,3,1\}$, $\{5,4,1\}$, $\{5,3,2\}$, $\{4,4,2\}$, $\{4,3,3\}$.
For each of these, you then compute the number of ways to arrange the elements.
e.g. for $\{8,1,1\}$, there are $\binom{10}{8}\cdot\binom{2}{1}\cdot\binom{1}{1}$ to arrange, e.g. eight a's, one b, and one c. You would need to multiply that result by $3$ to account for the fact that you could have eight of the other letters instead.
Similarly for $\{7,2,1\}$, you would do $\binom{10}{7}\cdot\binom{3}{2}\cdot\binom{1}{1}$, but this you would multiply by $3!=6$ because there are that many ways to determine the quantities of each letter you have.
And so on for each partition of $10$. It seems a bit painstaking to do all of it, so I won't, but hopefully you can take it from here, and maybe someone will come along with an easier method.
On
WLOG pick one element you choose to omit while creating that sequence. The number of possible sequences times 4 gives the final result because ${4\choose 1}={4\choose 3}=4$. Then consider that to generate a sequence containing each element at least once you can first fixate those 3 choices, you have ${10\choose 3}$ ways of doing so; for each of those ways you are left with $3^7$ choices to complete the series, because any of the 3 elements will be valid for any of the remaining 7 positions.
The total number of ways thus equals
${4\choose 1}{10\choose 3}3^7$
Edit: obviously the restraint of containing each element at least once leads to less ways than arbitrary sequences of length 10, which make for $3^{10}$, though obviously ${10\choose 3} > 3^3$ and thus I was mistaken for a moment. Instead, use inclusion-exclusion at that part of the proof.
On
First, select the $3$ elements to show up, in $\binom{4}{3}$ ways.
In all, there are $3^{10}$ sequences of $3$ elements, of which $\binom{3}{2} \cdot 2^{10}$ are of at most two symbols. Thus:
$$ \binom{4}{3} \left( 3^{10} - \binom{3}{2} \cdot 2^{10} \right) = 223908 $$
On
The Stirling number of the second kind $\left\{n \atop k\right\}$ counts the number of partitions of an $n$-set into exactly $k$ parts, so $k! \left\{n \atop k\right\}$ is the number of surjections of an $n$-set onto a $k$-set. Your desired count is hence $$\binom{4}{3} 3! \left\{10 \atop 3\right\} = 4 \cdot 6 \cdot 9330 = 223920.$$
If you choose $10$ items from a pool of $3$ elements, you of course have $3^{10}$ possibilities, although that will include some where not all are used.
If you choose $10$ items from a pool of $2$ elements, you have $2^{10}$ outcomes, although again not every option will have both elements present.
Finally if there is only one element to choose from, there's no choice and a single outcome. $1^{10}=1$.
To get to exactly three elements present, we can choose the elements $\binom 43 = 4$ ways and then use inclusion-exclusion to eliminate the deficient options:
$$\binom 43 \left[ 3^{10} - \binom 32 2^{10} + \binom 31 1^{10} \right]$$