no. of ways of forming a garland using $6$ red roses and $4$ white roses. The no. of ways that the garland can be made out of $6$ red and $4$ white roses so that all white roses not come together.
$\bf{My\; Solution::}$ First we will form a garlad using $6$ red roses, This can be done in $\displaystyle (6-1)!$ ways.
Now we will arrange $4$ white roses in $6$ gap produced by $6$ red roses, This can be done in $\displaystyle \binom{6}{4}\times 1$
because all roses are white.
So Total no. of ways is $ = \displaystyle 5! \times \binom{6}{4}$
But my answer is wrong.
please explain me in detail where my answer is wrong
Thanks
 A: This  problem  can  be  solved  with  the  Polya  Enumeration  Theorem
(PET). We assume that garlands that can be transformed into each other
by  rotations or  reflections are  considered the  same. 
Furthermore we do not treat the problem where the forbidden garlands are
precisely those that contain four white roses all in one block, 
as that is a trivial variation on the ordinary bracelet / necklace problem.
We  will solve the slightly more general problem of having $n$ red roses 
where $n\ge 4.$
First place  the four white roses  on the garland,  leaving some space
between them. These  spaces are where the red roses  go but there must
be at least one red rose in every one of the four spaces. The dihedral
group $D_4$ acts on these slots and therefore the answer is given by
$$[z^n] Z(D_4)\left(\frac{z}{1-z}\right).$$
We compute $Z(D_4)$ next. The rotations contribute
$$a_1^4+a_2^2+2a_4$$
and the reflections
$$2a_1^2 a_2 + 2 a_2^2$$
for a total result of
$$Z(D_4) = \frac{1}{8}
\left(a_1^4 + 3 a_2^2 + 2 a_1^2 a_2 + 2a_4\right).$$
The substituted cycle index becomes
$$Z(D_4)\left(\frac{z}{1-z}\right)
\\= \frac{1}{8}
\left(
\left(\frac{z}{1-z}\right)^4 
+ 3 \left(\frac{z^2}{1-z^2}\right)^2 
+ 2 \left(\frac{z}{1-z}\right)^2\left(\frac{z^2}{1-z^2}\right)
+ 2 \left(\frac{z^4}{1-z^4}\right)\right).$$
Introduce the predicate
$$q_m(n) = \begin{cases}
1 & \quad\text{if}\quad m|n \\
0 & \quad\text{otherwise.}
\end{cases}.$$
Extracting coefficients from the inner terms we get for the first term
$${n-4 +3\choose 3} = {n-1\choose 3}.$$
For the second term we get
$$q_2(n)\left(\frac{n}{2}-1\right).$$
For the third term we obtain 
(use e.g. the partial fraction decomposition)
$$\frac{1}{4} n^2 - n + \frac{7}{8} + \frac{1}{8} (-1)^n.$$
Finally the last term contributes $$q_4(n).$$
Collecting all contributions we get 
$$\frac{1}{48} n^3 - \frac{1}{16} n^2  - \frac{1}{48} n + \frac{3}{32}
+ \frac{1}{32} (-1)^n
+ \frac{3}{8} q_2(n)\left(\frac{n}{2}-1\right)
+ \frac{1}{4} q_4(n).$$
This yields the following sequence (starting at $n=1$):
$$0, 0, 0, 1, 1, 3, 4, 8, 10, 16, 20, 29, 35, 47, 56, 72, 84, 
104, 120, 145,\ldots$$
which is OEIS A005232.
In particular with six red beads there are $3$ garlands, which are
(first distribute four red roses into the spaces as that is the minimum,
leaving two red roses): 
one, the two red roses go into the same slot, 
two, the two red roses go into adjacent slots and 
three, the two red roses go into opposite slots.
A: First, do circular permutation of 6 red and thus we have 6 places for arranging 4 reds in six places linearly. Hence,
(6-1)!=120
and then after linear permutation
6P4=360
This can be clockwise and also anti-clockwise 
thus total types=120 * 360/2 =21600
