(a) There are 6 points in a plane. How many triangles can be formed with the points? (b) A family of 6 is to be seated in a row . In how many ways can this ...
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(a) There are 6 points in a plane. How many triangles can be formed with the points?
(b) A family of 6 is to be seated in a row . In how many ways can this be done if the father and mother are not to be seated together?
Answer Details
(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:
\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]
So 20 triangles can be formed.
(b) Seating 6 people in a row with father and mother NOT together.
Total arrangements of 6 people in a row: \(6!=720.\)
Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:
\[5!\times2!=120\times2=240.\]
Therefore arrangements with them NOT together:
\[720-240=480.\]
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