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952 Chapter 29 Statistics and probability distributions

Each combination ofr objects gives rise tor! permutations, but

P(n,r) =

n!

(n−r)!

Thenumberofcombinationsofndistinctobjects takenratatime is

( n

=

r)

P(n,r) n!

=

r! (n −r)!r!

Example29.16 Calculate the numberofcombinationsof

(a) sixdistinctobjects takenfive atatime

(b) ninedistinctobjectstaken nineatatime

(c) 25 distinctobjects takenfive atatime

Solution (a)

( 6

5)

= 6! ( ) 9

1!5! =6 (b) = 9! ( ) 25

9 0!9! =1 (c) = 25!

5 5!20! = 53130

( n

We can generalize the resultofExample 29.16(b)and statethat = 1.

n)

Example29.17 There are k identical objects and n compartments (n k). Each compartment can

hold only one object. In how many different ways can thekobjects be placed in then

compartments?

Solution The order in which the objects are placed is unimportant since all the objects are identical.

Placing the k objects is identical to selecting k of the n compartments (see

(

Figure29.7).Butthenumberofwaysofselectingkcompartmentsfromnisprecisely

. n

(

k)

n

Hence thekobjects can beplacedinthencompartmentsin different ways.

k)

n compartments

k objects

Figure29.7

Placingkobjectsinncompartments.

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