SOLUTION: 4. How many arrangements are there of Seven a’s, Eight b’s, Three c’s, Six d’s
with no occurrence of the consecutive pairs ca or cc?
I have tried by indirect
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Question 32081: 4. How many arrangements are there of Seven a’s, Eight b’s, Three c’s, Six d’s
with no occurrence of the consecutive pairs ca or cc?
I have tried by indirect approach. The total cases being 24!/(7!*8!*3!*6!)
The bad cases being when ca or cc are together. I am getting stuck here as I don’t know whether the c I take in case ca should be counted again in cc.
Thanks in advance as I have been working on this question for a long time.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
How many arrangements are there of Seven a’s, Eight b’s, Three c’s, Six d’s
with no occurrence of the consecutive pairs ca or cc?
I have tried by indirect approach. The total cases being 24!/(7!*8!*3!*6!)
The bad cases being when ca or cc are together. I am getting stuck here as I don’t know whether the c I take in case ca should be counted again in cc.
Thanks in advance as I have been working on this question for a long time.
KEY TO SOLVING THESE PROBLEMS IS TO DEVELOP A LOGICAL PROCEDURE WHICH
1.IS AMENABLE TO CALCULATION WITH KNOWN/STANDARD FORMULAE OR FROM FUNDAMENTALS
2.SATISFIES ALL POSITIVE REQUIREMENTS OF THE PROBLEM
3.EXCLUDES ALL NEGATIVE REQUIREMENTS OR EXCEPTIONS IN THE PROBLEM.
HERE THERE ARE CONSTRAINTS ON A AND C ...SO LET US LEAVE THEM ASIDE FIRST.
THERE ARE 8 B'S AND 6 D'=14 IN ALL...
THEY CAN BE ARRANGED IN 14P14/(8!6!)WAYS.=P WAYS................I
NOW THESE 14 LETTERS HAVE 13 INTER SPACES AND 2 OUTER SPACES..TOTAL 15
SINCE CC CAN'T BE TOGETHER EACH C HAS TO BE PLACED IN ONE OF THESE INTER SPACES
NUMBER OF WAYS FOR THIS =15P3/3!=Q WAYS.............................II
NOW IN EACH OF THIS CASE WE STILL HAVE 15-3=12 SPACES....SINCE CA (AND AC IS ALSO PROHIBITED?..FIRST LET US ASSUME SO AND DO.IF THAT IS NOT EXCLUDED YOU CAN MODIFY THE ANSWER FOR THAT TOO)IS ALSO PROHIBITED A'S HAVE TO COME IN THESE 12 SPACES ONLY.
BUT THERE IS NO RESTRICTION ON A'S BEING TOGETHER..SO WE CAN CONSIDER PLACING THESE 7 A'S IN 12 POSITIONS IN THE FOLLOWING CATEGORIES
LEGEND...7A MEANS 7A'S TOGETHER....
COMBINATIONS......NUMBER OF WAYS
7A....................... 12P1
6A+1A.................... 12P2
5A+2A.................... 12P2
5A+1A+1A................. 12P3/2!
4A+3A.................... 12P2
4A+2A+1A................. 12P3
4A+1A+1A+1A.............. 12P4/3!
3A+3A+1A................. 12P3/2!
3A+2A+2A................. 12P3/2!
3A+2A+1A+1A.............. 12P4/2!
3A+1A+1A+1A+1A........... 12P5/4!
2A+2A+2A+1A.............. 12P4/3!
2A+2A+1A+1A+1A........... 12P5/3!
2A+1A+1A+1A+1A+1A........ 12P6/5!
1A+1A+1A+1A+1A+1A+1A..... 12P7/7!
--------------------------------------------------------------
TOTAL NUMBER OF WAYS ..........................SUM OF ALL ABOVE=R WAYS....III
TOTAL NUMBER OF ARRANGEMENTS THEN =P*Q*R
NOW IF AC IS ALLOWED THEN WE SHALL HAVE IN ADDITION TO THE 12 SPACES MENTIONED ABOVE 3 MORE SPACES ALL TO THE LEFT OF C'S WILL BE AVAILABLE MAKING IT TOTAL 15 FREE SPACES,WHICH CAN BE PUT IN PLACE OF 12 ABOVE TO GET THE ANSWER.
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