Hello,
I'm confused and I don't know who is right.
Problem:
six children A,B,C,D,E,F are to be sitted together in a single row of six chairs. If A cannot sit next to B, how many different arrangements of six children are possible?
a) 240, b) 480, c) 540, d) 720, e) 840.
From what I read is that I should take 2 of the children as if it was 1, so then I will have 5! posibilities = 120
Then I take the 2 that are sitting together and find the possible sits for those 2; that equals to 2!=2
I multiply 120*2= 240 options.
Why in this problem do I have to substract 6!- 240 to equal 480 which is the right answer while in the following problem I don't have to?
Problem:
find the arrangements of 8 people if 3 must be sit together.
correct answer= 4320
I take 3 as a 1 person so I find 6!= 720, then I take the 3 that are sitting together and find the possible sits in this case 3!= 6
Then I multiply 720*6= 4320
I'm confused and I don't know who is right.
Problem:
six children A,B,C,D,E,F are to be sitted together in a single row of six chairs. If A cannot sit next to B, how many different arrangements of six children are possible?
a) 240, b) 480, c) 540, d) 720, e) 840.
From what I read is that I should take 2 of the children as if it was 1, so then I will have 5! posibilities = 120
Then I take the 2 that are sitting together and find the possible sits for those 2; that equals to 2!=2
I multiply 120*2= 240 options.
Why in this problem do I have to substract 6!- 240 to equal 480 which is the right answer while in the following problem I don't have to?
Problem:
find the arrangements of 8 people if 3 must be sit together.
correct answer= 4320
I take 3 as a 1 person so I find 6!= 720, then I take the 3 that are sitting together and find the possible sits in this case 3!= 6
Then I multiply 720*6= 4320
