"Order is allowed" means different permutations of the same objects are considered different.
For example, number of ways of picking 2 objects out of $\{a,b,c\}$ with "order allowed" will include $\{ab, ba, bc, cb, ac, ca\}$. A good way of understanding "order allowed but repetition not allowed" in your problem is to think you are picking the six objects one at a time from the ten objects. For instance, in my example $ab$ represents picking $a$ in the first pick and picking $b$ in the second.
The way to count with "order allowed but repetition not allowed" is to see how many choices you have at each pick and multiply them. For my example, I had three choices for the first pick viz. $a,b,c$ and for my second pick, i have two choices left. So the total number of ways is $3 \times 2 = 6$ which is the total number of possibilities in our enumeration viz. $\{ab, ba, bc, cb, ac, ca\}$
"Repetitions are allowed" generally means that an item can picked at another time, even though it has been picked once before.
A good way of understanding "order allowed and repetition allowed" in your problem is to think you are picking the six objects one at a time but you put the object back into the bag after you make note of what you picked. For instance, in my example $aa$ represents picking $a$ in the first pick and putting it back and then picking $a$ again in the second.
For example, number of ways of picking 2 objects out of $\{a,b,c\}$ with "order allowed and repetition allowed" will include $\{aa, bb, cc, ab, ba, bc, cb, ac, ca\}$. Do you see how $aa$ is now included? This represents the possibility of getting $a$ in both picks.
The way to count with "order allowed and repetition allowed" is to see how many choices you have at each pick and multiply them. For my example, I had three choices for the first pick viz. $a,b,c$ and for my second pick, I still have three choices. So the total number of ways is $3 \times 3 = 9$ which is the total number of possibilities in our enumeration viz. $\{aa, bb, cc, ab, ba, bc, cb, ac, ca\}$
If "Order is not allowed", then different permutations of the same objects are considered same.
For example, number of ways of picking 2 objects out of $\{a,b,c\}$ with "order not allowed and repetition not allowed" will include $\{ab, bc, ac\}$. Do you see how $ba$ is not included? This is because $ab$ has been included. The different orderings of $ab$ is irrelevant now.
A good way of understanding "order not allowed and repetition not allowed" in your problem is to think you are picking the six objects all at once. For instance, in my example $ab$ represents picking two objects in one go, where I picked $a$ and $b$, Clearly this experiment does not count permutations of $ab$ as different. All that matters is which object you received.
The way to count with "order not allowed and repetition not allowed" is to use $\binom{n}{r}$ formula. This represents the total number of ways of non-repeated picking of 'r' objects out $n$ objects in an unordered manner. So the total number of ways is $\binom{3}{2} = 3$ which is the total number of possibilities in our enumeration viz. $\{ab, bc, ac\}$
The trickiest case is the last one. This is the case where repetitions are allowed but order is not allowed.
For example, number of ways of picking 2 objects out of $\{a,b,c\}$ with "order not allowed and repetition allowed" will include $\{aa, ab, ac, bb, bc, cc\}$.I think no more explanation is needed for the above enumeration.
A good way of understanding "order not allowed and repetition allowed" in your problem is to think you are distributing $6$ identical objects in $10$ distinct urns. In my example, two identical objects have to be distributed in urns labeled $a,b,c$. For instance, $ab$ on the enumeration represents distributing one of my objects in an urn labeled $a$ and another object in an urn labeled $b$.
The way to count with "order not allowed and repetition allowed" is to use $\binom{n + r - 1}{r}$ formula. This represents the total number of non-negative solutions to the equation $x_1 + x_2 + \cdots + x_r = n$. So the total number of ways is $\binom{3+2-1}{2} = 6$ which is the total number of possibilities in our enumeration viz. $\{aa, bb,cc, ab, bc, ac\}$