Permutaions questions...? PLEASE HELP...?

Question

1) Amy, Brian, Cheryl, Danny and Eric went to a concert. How many arrangements are possible when they sit in five adjacent seats if Brian refuses to sit next to Danny?


2) At an art exhibition 7 paintings are to be hung in a row along one wall. Given that 3 paintings are by the same artist, find the number of arrangments in which any one of these paintings is hung at the begining of the row but neither of the other 2 is hung at the end of the row.

Please explain clearly.

Answer 1

The prior two answers for problem 1) are incorrect.
There are 2! = 2 ways to arrange Brian and Danny, and 3! = 6 ways to arrange the remaining people. There are 5! = 60 ways of arranging everyone without the constraint.

Therefore, the number of ways of arranging everyone without Brian and Danny sitting together is 60 -12 = 48 ways! Here they are:
ABDCE
ABDEC
ABEDC
ABECD
ACDBE
ACDEB
ADBEC
ADCEB

BACDE
BACED
BADCE
BADEC
BAECD
BAEDC
BEACD
BEADC
BECAD
BECDA
BEDAC
BEDCA

CABDE
CABED
CADBE
CADEB
CDABE
CDAEB
CDBAE
CDBEA
CEABD
CEADB
CEBAD
CEBDA

DABEC
DACEB
DBACE
DBAEC
DCABE
DCAEB
DEBAC
DECAB

EABDC
EACDB
EBACD
EBDCA
ECABD
ECDAB
EDBAC
EDCAB

Hope this helps!

Answer 2

1. there are 5 seats.

first,fix brian and danny in any 2 places such tht they arent adjacent.in 2!ways.

the remaining 3 can sit in 3! ways.

so, total no of arrangements = 2! x 3! = 2 X 6 = 12 ways.

2. there are 7 paintings..

first,fix the 3 paintings in the req way(one a beginning and other 2 in between) in 3! ways.
now the other 4 can be arranged among themselves in 4! ways.

total no. of ways of arranging=3!4!=6 x 24 = 144 ways

hope u r satisfied.

Answer 3

there are 5! combination( i.e. 60 ) with out the condition
and 2*4! possibilities for brian sitting next to danny. ie is 48
so possibilities of brian not sitting next to danny is 60-48 =12

likewise do the other one