fr743 wrote:GMATGuruNY wrote:ruplun wrote:If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together?
Good = Total - Bad.
Total:
Number of ways to arrange the 5 children = 5! = 120.
Bad:
In a bad arrangement, the 2 siblings sit adjacent to each other.
Let [AB] represent the 2 siblings sitting adjacent to each other.
Number of ways to arrange the 4 elements [AB] and the 3 other children = 4! = 24.
Since AB can be reversed to BA, we multiply by 2:
24*2 = 48.
Good = 120-48 = 72.
OABAA
OAABA
OAAAB
AOABA
AOAAB
AAOAB
BAOAA
BAAOA
ABAOA
BAAAO
ABAAO
AABAO
12 => MULTIPLY BY 2 Total 24.
Why am I wrong?
I'm not sure why you have 3 A's in your arrangements.
Call the elements {AB], C, D and E and count all the possible arrangements:
[AB]CDE
[AB]CED
[AB]DCE
[AB]DEC
[AB]ECD
[AB]EDC
There are 6 arrangements in which A occupies position 1 and B occupies position 2.
Since [AB] can be reversed to [BA], we multiply by 2:
Number of of arrangements in which [AB] occupies the first 2 positions = 2*6 = 12.
[AB] can occupy 4 sets of adjacent positions:
Positions 1 and 2
Positions 2 and 3
Positions 3 and 4
Positions 4 and 5
Each set of adjacent positions will yield 12 possible arrangements, as shown above.
Since there are 4 sets of adjacent positions, we multiply by 4:
Total arrangements in which [AB] sit adjacent to each other = 12*4 = 48.
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