How many different arrangements are possible if $3$ letters are randomly selected from the word CHALLENGE and arranged into ‘words’?

Question

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 $246$.

Answer 1

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 $246$.

Answer 2

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.