permutation

Question

for aaaabbbcccdde find no of permutation such that

1)no two c are together

2)no 3 c are consecutive

Answer 1

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

Answer 2

1.  No consecutive c's  ,

There are  (N-4)*(N-3)/2 ways to select non consecutive positions for c's so number of way to place c's is (13-4)*(13-3)/2 = 45

There are 10!/(2!*3!*4!) = 12600 ways to place other letters in remaining places , so total number of permutation = 45*12600 = 567000

2. No 3 c's are together ,  permutation for all c's together is 11!/2!3!4! = 138600 and the possible permutations are = 13!/(2!3!4!3!) = 3603600  . So answer  = 3603600-138600 = 3465000