Finding the number of counters having each colour. There are counters available in $7$ different colours. Counters are all alike except for the colour and they are atleast $10$ of each colour. Find the number of ways in which an arrangement of $10$ counters can be made. How many of these will have counters of each colour.  
As there are ten places to fill and each can be filled with $7$ colours the number of arrangement $7^{10}$.  
For the second question I note that as there are atleast $10$ of each we have cases ranging from all $10$ of same colours. How to handle all these taking each case and then finding perputations seems to be a tiring job. 
 A: Your $7^{10}$ is correct. If you have all colors represented, you have a limited number of combinations of colors.  You can have four of one color and one of each other, or ...  Now compute the number of ways of getting each of these.
A: Hint for part two
Suppose you have chosen $3$ of one color, $2$ of another, and $1$ each of the rest,
viz. a $\;\;3-2-1-1-1-1-1$ of a kind, then there will be:
$\left[\binom71\binom61\;\text{ways to choose the colors}\right]\times\left[\frac{10!}{3!2!}\text{ ways to place them}\right]$
Work out similarly for other patterns and add up.
A: Your answer for the first question is correct.
For the second question, we can use the Inclusion-Exclusion Principle.  
We must exclude those solutions in which there are fewer than seven different colors used.  
There are $\binom{7}{1}$ ways of excluding one of the seven colors and $6^{10}$ ways of arranging $10$ counters using the six remaining colors.
There are $\binom{7}{2}$ ways of excluding two of the seven colors and $5^{10}$ ways of arranging $10$ counters using the five remaining colors.
There are $\binom{7}{3}$ ways of excluding three of the seven colors and $4^{10}$ ways of arranging $10$ counters using the four remaining colors.  
In general, there are $\binom{7}{k}$ ways of excluding $k$ of the seven colors and $(7 - k)^{10}$ ways of arranging $10$ counters using the $k$ remaining colors.  
By the Inclusion-Exclusion Principle, the number of arrangements in which each of the seven colors is used is 
$$\sum_{k = 0}^{7} (-1)^k\binom{7}{k}(7 - k)^{10}$$ 
