1) no two 'c' are together.
X : represents places where we can put 'c'.
X a X a X a X a X b X b X b X d X d X e X
Required no.of ways = no.of arranging 4 a's , 3b's ,2 d's and 1 e's and then selecting 3 places out of 11 places and putting 3 c's there.
= ((10!) /( 4! * 3! * 2! )) * (11C3 * 3!/3! ) = 2079000
2) Here required no of ways = Total no of ways of arranging 13 letters - No of ways in which all 3 c's are together(i.e consider all 3 c's as one unit ,thus arrange 11 letters and finally arrange 3 c's among themselves
= ((13!) /( 4! * 3! * 3! * 2! )) - ((11!)/( 4! * 3! * 2! )) * (3!/3!) = 3603600 - 138600 = 3465000