Cool Combinations!

Without accounting for the 2 R's being together or not, there are 7!/(2!*2!) different ways of arranging the letters (divide by 2! for each duplicated letters).

Then we remove all bad combinations (those with the 2 R's together). We consider the 2 R's as 1 entity. There are 6!/2! ways.

So the total is: 7!/(2!*2!) - 6!/2! = 900.

because the two 'R's are never together, thus, there are 15 ways of arranging the two 'R's into two positions. Then, there are 10 ways of arranging the two 'A's into two position. Finally, there are 3x2x1 = 6 ways of arranging three letters: N, G, E. So there are 15 x 10 x 6 = 900 ways

We can arrange the letters in 7!/(2! 2!) ... since total letters are 7 and letters R & A are repeated twice , so we divide 7! by 2! 2!. Then we consider the ways in which we can arrange the word in such a way that 2 R's are together . This is possible in 6!/2! ways . So now subtracting the ways of 2 R's being together from the ways in which we can arrange the letters without considering any condition , we get the ways in which we can arrange the letters such that 2R's are not together . So , 7!/(2!*2!) - 6!/2! = 900. is the ans . Simple

$\frac{7!}{2!2!} - \frac{6!}{2!}=$

$1260-360=$

$\boxed{900}$

7!/(2!)(2!)-6!/2!

7!/(2!*2!)-(6!/2!)

7!/(2!2!) - 6!/2!=900