+28 In how many ways can three pairs of siblings from different families be seated in two rows of three chairs, if siblings may not sit next to each other in the same row?
pls help me as quick as possible
+6 Okay. Let's consider the siblings as A1 and A2, B1 and B2, C1 and C2. SInce they are sitting in two rows we will fill the first row in first. So A1 is in the first, B1 is in the second, and C1 is in the third. Now we move on to the second row which is A2, B2, and C2.
However, there is another possible arrangement for the first row. That is A2 for the first, B1 for the second, and C1 for the third. Then the second row is A1, B2, and C2.
These are the only two possible arrangments that will satisy the conditions given. Thus, there are two ways to seat three siblings so that siblings don't sit next to each other.
+743 To solve this, let's first consider the case where there are no restrictions on the seating arrangements. In this case, there are 6! ways to arrange the 6 siblings in the 6 chairs.
Next, let's consider the number of arrangements where siblings are seated next to each other. For each pair of siblings, there are 2 ways to arrange them (either sibling can sit in the front or back chair). Since there are 3 pairs of siblings, there are 23=8 ways to arrange the siblings in pairs.
However, we have double-counted the arrangements where two pairs of siblings are seated next to each other. For each arrangement of two pairs of siblings, there are 2 ways to arrange the third pair. Therefore, we have overcounted the number of arrangements with two pairs of siblings by 8⋅2=16.
Finally, let's consider the number of arrangements where all three pairs of siblings are seated next to each other. There are 2 ways to arrange the three pairs of siblings (either all three pairs can be in the front row or all three pairs can be in the back row). Once the pairs are arranged, there are 2 ways to arrange each pair within their respective row. Therefore, there are 2⋅23=16 arrangements where all three pairs of siblings are seated next to each other.
Therefore, the total number of arrangements where siblings cannot be seated next to each other is 6!−8⋅2−16=552.
