Arrangement of words $\mathrm{CALCULATE}$ such that each word starts and ends with a consonant. The number of words that can be arranged with the letters of the word $\bf{CALCULATE}$
such that each word starts and ends with a consonant is
$\underline{\bf{My\; Try}}::$ Given Letter has $\bf{2A\;,2C\;,2L\;,U,T,E}$. Now here Vowels $\bf{=2A\;,2L\;,E,U}$
and Constant $\bf{ = 2C\;,T}$. Now we have to form a no. which start and end with constant::
So we have Two Different Possibilities::
$\bf{\bullet}$ If First and Last position Contain same Letter:
$\bf{\boxed{C}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{C}}$
Now we have to enter $\bf{2A\;,2L\;,U,E,T}$ in  these Boxes containing $\bf{+}$ sign.
So Total no. of ways $\displaystyle \bf{ = \frac{7!}{2! \times 2!}}$
$\bf{\bullet}$ If First and Last position Contain Different Letter:
$\bf{\boxed{C}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{T}}$ OR $\bf{\boxed{T}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{+}\boxed{C}}$ 
Now we have to enter $\bf{2A\;,2L\;,U,E,C}$ in  these Boxes containing $\bf{+}$ sign.
So Total no. of ways $\displaystyle \bf{ = \frac{7!}{2! \times 2!}+\frac{7!}{2! \times 2!}}$
So Total no. of ways in which Letters start with Constants is $\displaystyle = 3 \times \frac{7!}{2! \times 2!}$
Is My solution right? If not how can I solve it?
 A: In the usual classification, L is a consonant. On the assumption it is a vowel, the cases division that you made is correct. The approach could be modified to deal with the fact that L is a consonant. It would become somewhat more complicated. 
We solve the problem with L a consonant, and without dividing into cases. Put ID's on the letters that are the same, to make them distinct. So we have $9$ different letters, of which $5$ are consonants.
The first slot can be filled with a consonant in $5$ ways. For each such way, the last slot can be filled in $4$ ways. And then the "middle" can be filled in $7!$ ways, for a total of $(5)(4)(7!)$.
Now take off the ID numbers. When we take off the ID's on the A's, the number of distinct words gets divided by $2!$. There is also division by $2!$ when we remove the ID's from the C's, and from the L's. We end up with 
$$\frac{(5)(4)(7!)}{(2!)^3}$$
distinct words that satisfy the specification.
A: CALCULATE
No of Consonants = 2C + 2L + 1T =5 
No of vowels = 2A+ 1U +1E= 4
At first place ,no of letters that can be arranged =5 
At last place ,no of letters that can be arranged=5-1=4 (as one letter is used at first place)
Now 7 places are there between these 2.
If we consider no repetition,remaining consonants are 3(2 used at first and last place)+no of vowels 4=7 letters are available to arrange in these 7 places and they are arranged by 7!
1 2 3 4 5 6 7 8 9  no of places
5 7 6 5 4 3 2 1 4 no of possible arrangements
Now total ways= 5*4*7!/2! *2! *2!,,...(Here C A and L are repeated twice and hence should be divided)=5*7!/2
