Combinatorics

[knight247]

There are eight orators A,B,C,D,E,F.G & H. How mant ways can they speak at a function if C has speak before A and A has to speak before D and D has to speak before H though not necessarily right after each other.

(A)40320
(B)20160
(C)1680
(D)3360
(E)6720

Don't have an OA on this. Sorry. Detailed explanations would be appreciated.

[cans]

c,a,d,h are to be in order c->a->d->h.
c,a,d,h can be arranged in 24 ways (4!)
out of these 4! we need only 1.
Thus 8!/4! = 1680
IMO C

[prashant.mishra]

Eight people can be arranged in 8! ways. C|A|D|H can't arrange in themselves.

Thus the number of ways=

8!/4! ( to negate the arrangements made by C, A, D and H, we need to divide the total arrangements by their internal arrangements ie. 4!)

= 1680 Ans.

IMO C.

[knight247]

Ok here is where I have an issue. Does the 1680 ways include a sequence like the following
C B A E D F H. This is what I'm not quite able to follow. Hope someone can clarify this for me. Thanks

[prashant.mishra]

Yes. The 1680 ways includes this sequence and all sequences in which C speaks before A, A before D, and D before F. One must understand that to cancel the arrangements among a particular set of people, DIVIDE the total arrangements possible by the number of arrangements of the members whose ORDER should be fixed. In case you are finding difficulty still understanding this question, take a smaller sample. lets say for 2 people A & B. If they were to arrange themselves where anyone could speak first, there would be 2! ways. Now if we say, only A speaks first then we have 2!/2! =1 way only.

Take 3 guys now. A, B & C.

Toatal ways of arrangement: 3! They are: ABC, ACB, BAC, BCA, CAB, CBA. Now we say A is before C. How many ways? 3!/2! = 2 ways. They are : ABC, BAC, ACB.

Take 4 now:

A, B, C, D. Total ways: 4!=24 They are:
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCDA, BCAD, BDCA, BDAC,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA.

Lets say A will always speak before B . How many ways? 4!/2!=12 They are:

ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
CABD, CADB, CDAB,
DABC, DACB, DCAB.

Lets say A will speak before B and B will speak before C. How many ways? 4!/3!=4 They are:

ABCD, ABDC, ADBC,
DABC.

Boy, I can't explain any further...need some water !!

But hope this helps !!

Ok here is where I have an issue. Does the 1680 ways include a sequence like the following
C B A E D F H. This is what I'm not quite able to follow. Hope someone can clarify this for me. Thanks

[knight247]

Thanks for the detailed explanation Prashant. Appreciate it!!