HP 10S Manuel
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HP 10s Probability – Rearranging Items
Rearranging Items
Practice Solving Problems Involving Factorials,
Permutations, and Combinations
hp calculators
HP 10s Probability – Rearranging Items
hp calculators - 2 - HP 10s Probability – Rearranging Items - Version 1.0
Rearranging items
The calculation of probabilities often involves counting the ways in which items can be rearranged. We have three
functions at our disposal for this purpose: factorial, permutations and combinations.
If nis a positive integer, its factorial (whose symbol is
!n
) is the product of all the positive integers up to and including n:
1221 !!!"!"!=...)n()n(n!n
0! is defined as 1.
A permutation of rfrom n is a way in which a set of relements may be chosen in order from a set of nelements. In other
words, it’s an ordered subset of a set of distinct objects. The number of possible permutations, each containing robjects,
that can be formed from a collection of n distinct objects is given by:
!
nP
r=P
r
n=nP
r=P(n,r)=(n)r=n!
(n"r)!
The above expression also shows the multiple ways of symbolizing permutations. When a permutation involves all the
elements of a set (i.e. r = n) then it is called a rearrangement, and also as shuffle especially when cards are used. Notice
that in this case:
!n
!
!n
PP n
n
n
r=== 0
A combination is a selection of one or more of a set of distinct objects without regard to order. These are the different
ways of denoting combinations, and the number of possible combinations, each containing robjects, that can be formed
from a collection of ndistinct objects:
!
nCr=Cr
n=nCr=C(n,r)=(r
n)=n!
(n"r)!r!
As far as rearranging items is concerned, whether order is important or not is the only difference between combinations
and permutations.
One of the most important theorems on combinations is Vandermonde’s theorem, which states that:
11
1
!!
!+= n
r
n
r
n
rCCC
The factorial function was traditionally used for calculating permutations and combinations, which show up in many
discrete probability distribution calculations, such as the binomial and hypergeometric distributions.
On the HP 10s all these functions are executed using the b,F and c keys. The HP 10s errors for factorials of
integers greater than 69, but this covers most of the problems you will probably find, although numbers larger than 69
can be used for combinations and permutations, if the result does not cause an overflow.
hp calculators
HP 10s Probability – Rearranging Items
hp calculators - 3 - HP 10s Probability – Rearranging Items - Version 1.0
Practice solving problems involving factorials, permutations, and combinations
Example 1: Calculate the number of combinations and permutations of 15 objects taken 5 at a time.
Solution: The number of combinations is calculated by first keying in the value of n (it can be the result of a
calculation) then pressing Fand finally keying the value of r. =displays the result:
15 F5=
Likewise, the number of permutations is calculated by pressing:
15 Ac 5=
Answer: 3003 and 360360.
Example 2: Calculate the number of ways that six people can line up for a photograph.
Solution: Since the order does matter in this example, the problem is solved as a permutation:
6
6
P
. Press:
6 Ac 6=
Note that
1234566
6
6!!!!!== !P
, so this example can be solved using the factorial function,
which is a postfix function on the HP 10s:
6 Ab
The =key is not required now, because the result of a postfix function is automatically displayed as soon
as its key is pressed.
Answer: 720
Example 3: In how many different ways can five seats be filled by a group of ten persons?
Solution: Again, the order is important ,so the answer is
10
5
P
. Press
10 Ac5 =
Answer: 30,240
Example 4: How many different hands of 5 cards could be dealt from a standard deck of 52 cards? Assume the
order of the cards in the hand does not matter.
Solution: Since the order of the cards in the hand does not matter, the problem is solved as a combination:
52
5
C
52 F 5=
hp calculators
HP 10s Probability – Rearranging Items
hp calculators - 4 - HP 10s Probability – Rearranging Items - Version 1.0
Answer: 2,598,960 different hands.
Example 5: If five cards are dealt from a standard deck of 52 cards, calculate the probability of these five cards
containing four-of-a-kind.
Solution: The probability is the number of favorable events divided by the number of possible events. There are 13
ranks of four-of-a-kind, the fifth card being any of the remaining 48 cards. The number of possible five-card
hands has been calculated in the previous example: it is given by the combination
52
5
C
. Therefore, the
probability of being dealt four-of-a-kind is:
!
P=possible four - of - a - kind hands
possible five - card hands =13 "48
(5
52 )
The keystroke sequence is:
13*48/52 F 5=
Answer: 0.000240096. Approximately, 1 out of 4165 hands (which is calculated by inverting the result).
Example 6: Find the probability that at least two of 23 people have the same birthday.
Solution: The probability that at least two of npersons have the same birthday is given by:
!
1"P
n
365
365n
. In this
example n = 23:
1-365 Ac 23/365L23=
Answer: 0.507297234. The probability is greater than 50%. You can easily verify that for n ≤22, the probability is
smaller than 50%, so 23 is the smallest number of persons such that the probability of at least two of them
having the same birthday is greater than 50%!
Example 7: Evaluate the binomial density function f(x) at x=4 for a constant probability of success (p) of 0.49 and 6
Bernoulli trials (n).
Solution:
xnxn
x)p(p)()x(f !
!""=1
. In our case:
4646
449014904 !
!""=).(.)()(f
6 F4*.49 L4*W1-.49XLW4-2X=
Answer: f(4) = 0.224913711
Example 8: If you flip a coin 10 times, what is the probability that it comes up tails exactly 4 times?
Solution: It is an application of the previous example. In this case, n = 10, x = 4 and p = 0.5:
10 F 4*.5 L 4*.5LW10-4X=
Answer: 0.205078125. If you flip a coin 10 times, there is a 20.51% chance of seeing heads 4 times.
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