How many permutations of the letters in the word MISSISSIPPI are palindromes? How many of permutation of letters in  the word "$\bf{MISSISSIPPI}$" are palindromes?
$\text{My Try}:$ Palindrome is a word which read as  same forwards and backwards.
Here word "$\bf{MISSISSIPPI}$" contain $\bf{4I\;,4S\;,2P}$ and $\bf{1M}$. So total $11$ letters.
So must be in the form of $-----M-----$,
so we put $\bf{2I\;,2S\;,1P}$ on left of $M$,
which can be done as $\displaystyle =\frac{5!}{2!\times 2!\times 1} = 30$.
But I did not understand how can we not permute $\bf{2I\;,2S\;,1P}$ on right of $M$?
 A: We permute the letters on the right in the way which will make the $11$ letter word a palindrome. 
For example, we begin with $${\bf-----M-----}.$$ There are $30$ ways of permuting the two $\bf{I}$'s, two $\bf{S}$'s, and one $\bf{P}$ that we must use on the left. Suppose we choose the string $\bf{IPSSI}$ so that we have $${\bf IPSSIM-----}.$$ As we want the word to be a palindrome, we need to use the string $\bf{ISSPI}$ on the right, which is just the string we used on the left but reversed. In this case we get $${\bf IPSSIMISSPI}.$$
In general, we don't need to consider the permutations of letters on the right because the string is fixed once the string on the left is chosen, so the number of palindromes is just the number of choices for the string of letters on the left, which is $30$.
A: Once you put the letters to the left of $M$ then the letters to the right are fixed. They are the letters to the left in the reverse order
A: You figured out the placement of the M nicely. Once we know that, we need only consider the left side since once we place a letter on the left side, the same letter must be placed on the corresponding place on the right side of M. 
Then there are $5$ spots with $5$ possible letters to choose from, that is $5!$ total combinations. However, both the I and S repeat (since we have $2$ of them to place on the left side of M), to avoid repetitions, we must have
$$
\frac{5!}{2! \cdot 2!}=\frac{120}{4}=30
$$
So there are $30$ palindromes. 
A: When you fix a letter on the left side lets say P at the 1st position, the the last position is fixed with the letter P as the word should be a palindrome.
So only one side of the letter M can be permuted and the other side gets fixed automatically the condition being the word is a Palindrome !
So possible ways of arranging on one side : $$\frac{5!}{2! \cdot 2! \cdot 1!}=\frac{120}{4}=30$$
