Question is described here last question G1†
Solution is here last solution
Please explain the solution in simple language.
And please explain how to do it using partition.
My approach using partition:
$r: red\ beens,\ o: orange\ beens,\ y: yellow\ beens,\ g: green\ beens,\ b_l: blue\ beens,\ b_k: black\ beens,\ w: white\ beens,\ v: voilet\ beens$
$r = o,\quad y = g,\quad b_l = b_k - 1,\quad w = v - 3$
$r + o + y + g + b_l + b_k + w + v = 200$
by given info:
$r + y + b_k + v = 102$
Using Partition rule:
If we partition n into k parts: $n_1,n_2,....,n_k$
$$ no.\ of\ ways\ we\ can\ do\ it= \frac{n!}{(n-(n_1 + n_2+...+n_k))*n_1!*n_2!*...*n_k!} $$
$$if \quad n_1 + n_2 + ....+n_k = n$$
$$ no.\ of\ ways\ we\ can\ do\ it= \frac{n!}{n_1!*n_2!*...*n_k!} $$
now our n = 102, k = 4, $\quad n_1 = r,\ n_2= y,\ n_3=b_k,\ n_4=v$
$$number\ of\ jars = \frac{102!}{r!*y!*b_k!*v} $$
now am stuck at this point where am lacking or complete approach to this question is wrong ?
This is a stars and bars problem. The way you have done it, you want the number of integer solutions to $$r + y + b_k + v = 102$$ subject to $$\begin{align}r&\geq0\\y&\geq0\\b_k&\geq1\\v&\geq3\end{align}$$ So, put $1$ bean in the black jar, $3$ beans in the violet jar, and distribute the remaining $98$ beans in the $4$ jars, This gives $${98+4-1\choose4-1}={101\choose3}$$ ways.
EDIT
The approach with partitions is not fruitful, because you'd actually have to know what the partitions are. In those vases where the partitions are distinct, you can assign them to colors in $24$ ways, but if two are the same there are only $12$ assignments, and so on. This doesn't even address the complication that only certain values are acceptable for black and violet.