How many different arrangements are possible if $3$ letters are randomly selected from the word CHALLENGE and arranged into ‘words’? How many different arrangements are possible if $3$ letters are randomly selected from the word CHALLENGE and arranged into ‘words’?
$$\frac{9P3}{2! \cdot 2!} = 126$$
but the answer is $246$. 
 A: Consider the words with all different letters: select from CHALENG:
$${7\choose 3}\cdot 3!=210.$$
Consider the words with two letters L: select from CHAENG:
$${6\choose 1}{3\choose 1}=18.$$
Consider the words with two letters E: select from CHALNG:
$${6\choose 1}{3\choose 1}=18.$$
Add up to get $246$.
A: The given word is
$$CHALLENGE $$
we can write it as $$CHALLEENG$$
where there are total 9 words and 7 distinct words
We can think of this problem as- there are 9 balls in a box A out of which there are 7 balls of different colur and two balls which share the same colour out of 7 balls.
{R,B,G,Y,V,I,O,R,G}.Now in how many ways can we take out 3 balls and keep it in another box B ?  
CASE $1$
We can keep balls of all different colours in the box B.     
.ie 
$$7P3$$
$${7\choose 3}\cdot 3!=210 ways.$$
CASE $2$
Say box B has 3 different holes in which 3 balls can be kept 
We can keep 2 balls of same colour different colours in two holes and in the 3rd hole we can keep 6 different balls.
Therefore in the third hole 6 different balls can be kept.which can we again be done in three places 
.ie 
$$6*3$$
$$=18 ways.$$
Now case two repeates itself of the ball of another colour therefore again 18 ways
therefore total number of ways of doing so-$$18+18+210=246ways.$$
PS-I know that the solution to the problem was already given ,but i thought of doing it in another way.
