Page 169 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Definition: A permutation of n objects taken r at a time is a selection of r objects from a total of
n objects (r  n), where order matters.

Example: r = n: Suppose you have 6 (distinguishable) people to seat in 6 chairs. The first
person may choose her/his chair from 6 possibilities. The second person must choose from the
5 remaining chairs, etc., until the 6 person has to take the only chair that’s left. This arrangement
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can be done in 6×5×4×3×2×1, or 6! (“six factorial”) ways. Although this problem is somewhat
ambiguous, assume that each of the 14 objects, not each type of object, is a different color and
therefore distinguishable from the others.)

Example: r < n: Now suppose that you have 6 (distinguishable) people but only 4 chairs;
so only 4 of the 6 people will be able to sit down. How many arrangements of the 6 people can
be seated in the 4 chairs? This arrangement can be done in 6 × 5 × 4 × 3 ways.

Formulas: The general formula for a permutation of n objects arranged r at a time is P(n, r) = n!/
(n – r)!
Note that in Ex. P1, r = n, so n!/(n – r)! = n!/(n – n)! = n!/0! Since 0! is defined to be 1, we have n!/
1 = n!. Therefore, the answer to P1 is simply 6!.
Applying the formula to Ex. P2 yields 6!/(6 – 4)! = 6!/2! = (6 × 5 × 4 × 3 × 2 × 1)/(2 × 1) = 6 × 5 × 4 × 3.
Thus, the general formula applies in both cases, whether r = n or r < n.

Example: Suppose that you have 3 objects, but they are not all distinguishable. For
example, suppose that you have 1 red square and 2 green triangles. How many different
arrangements of those 3 objects are possible?

If the two green triangles switched places with each other, would the arrangements be
distinguishable from the ones given above? No, they would not. Therefore, the 2! ways the green
triangles could be rearranged must be eliminated. So if we have 3 objects, where 2 objects are the
same, then the number of possible arrangements is P(3, 3) = 3!/2! = (3 × 2 × 1)/(2 × 1) = 3 (not 3!).

Here objects of the same type are the same color and therefore not distinguishable from each
other.

Example: Suppose that you have 6 objects—1 red square, 2 green triangles, and 3 yellow
hexagons. How many different arrangements of the 6 objects are possible? In this case, you
have 2 indistinguishable green triangles and 3 indistinguishable yellow hexagons. Note that
you have 2! ways to interchange the 2 triangles and 3! ways to interchange the 3 hexagons. So
you must take those out of the total possible permutations.

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