

Phone Number Permutations
Name: William F.
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: N/A
Question:
What are the odds against two seven digit phone numbers being permutations of each
other? More precisely, if neither phone number is specified in advance? And more precisely
still, in the early 70's? Back then phone numbers never had a zero or 1 in the first three
digits (the exchange). This is I think harder than it looks, because you have to take into
account that with repetitions there are going to be fewer than 7! permutations of any randomly
selected seven digits.
Replies:
This is a 2part problem of the number of permutations of digits WITH REPLACEMENT ALLOWED, that is, any digit can appear more than one time. Consider a telephone "number" as: "abcdefg" the three digit exchange 'abc' and the four digit number 'defg'.
Most generally: any of the digits:a,b,....,f,g could be 0,1,...,8,9. However, since you are back in the 70's the first three digits:
'a,b,c' are restricted to the selection of the 8 digits 2,3,...,8,9 WITH REPLACEMENT ALLOWED. So that '222' is a perfectly allowed exchange. There are then 8*8*8 =512 possible choices for these three digits 'a,b,c'.
Each of the four remaining numbers: d,e,f,g can be 0,1,2,...,8,9 or ten possible values for each digit d,e,f,g. So there can be 10*10*10*10= 10^4 possible choices since REPLACEMENT is allowed. So, for example, d,e,f,g could be 3333 is O.K.
Now the 512 possible values of 'a,b,c' can be matched with any of the 10^4 possible choices of d,e,f,g. So together there are 512*10^4 = 5.12*10^6 possible phone numbers.
Assuming that each possibility is equally allowed, each number has an equal probability of:
1/[5.12x10^6] = 1.95x10^7 which is about 2 chances in 10million, or 200
parts per billion.
In the "real world" the number is smaller because there are other constraints that further DISALLOW certain combinations of either 'a,b,c' or 'd,e,f,g'. For example all: 911d,e,f,g numbers are probably reserved, and all d,e,f,g = 0000 are also reserved. The more recent addition of "area codes" adds another 512*(5.12x10^6) or ~2.62x10^9 possible numbers if the same constraints apply to area codes as apply to exchange numbers.
ZIP codes are also an example. For a 5digit ZIP code with no restrictions and REPLACEMENT IS ALLOWED there are 10^5 possible locations, and with the addition of the additional 4digit extensions gives 10^9 possibilities. It is evident why license plates in many states are combinations of letters and numbers because the alphabet is a base  26 alphanumeric. Of course, here too some letter combinations are reserved or disallowed for various obvious reasons that some combinations might be
socially unacceptable.
Vince Calder
FOLLOW UP QUESTION:
I'm still puzzled about the second part of
the question though: what are the odds against a random sevendigit phone
number containing the same digits in the same quantities as another one?
Obviously if my number is 7777777 no one else will have a permutation
of that number. Whereas if my number is 7654321, there are 7!1 other
permutations available. So the repeating digits is what makes the
problem hard. My thought was to try to figure out the number of repeating digits
in the average phone number, and take it from there. Any ideas? WF
The "simple" task of counting is often not quite so simple. The reason? Often it is not clear "what" is
being counted, and there are not any straightforward, plugin algorithms for determining "how" to count
"what" is being counted. The approach used in resources on "counting" seems to be to begin with various
"examples" of counting problems  population sizes, permutations, combinations, probabilities  and then,
in one way or another, say to the student/reader, "Now you see how this goes, try these
examples/problems." Consequently, each new counting task becomes a new problem. One filter that can be
employed to help with this ad hoc approach is to ask the questions, "What is the sample space being
counted?" and "Is what is being counted involve the entire sample space, or does it involve some
identifiable subset of the sample space?" If the "counting" involves the whole sample space, then the
"counting" process involves either multiplicative possibilities or additive possibilities. If the
"counting" involves some identifiable subset of the sample space, then the "counting" process involves
permutations and/or combinations that identify elements of the sample space that all have some property
in common.
A second filter that can be employed to assist in the ad hoc process is the question, "Does, or does not,
the "counting" process remove elements of the sample space from further consideration?" If the "counting"
removes elements of the sample space from further consideration, then the counting involves permutations
and/or combinations. If the "counting" process does not remove elements of the sample space from further
consideration, then the "counting" process involves all of the elements, all of the time.
The "probablility" of an event "P" is, by definition, the reciprocal of the number of elements of the
entire set, or of an identifiable subset, call it "N", that have some common property. That is P=1/N.
The present question can be separated into two parts: 1. How many 3digit telephone exchanges, abc, are
there if the integers a, b, and c can EACH have one of the eight values: 2, 3, 4, ..., 9 with no further
restrictions. The signal word here is EACH. That signals that the selection of 'a' or 'b' or 'c' does NOT
exclude any other elements from further consideration in the counting process. So 'a' can have eight
values: 2, 3, ..., 9. And 'b' and 'c' can have the same eight values. So each of the eight values of 'a'
can be followed by eight values of 'b', and each of those eight values of 'b' can be followed by eight
values of 'c'. The total number of possible 3digit exchanges is then: 8 x 8 x 8 = 512. The counting
process is: 222, 223, 224, ..., 748, ..., 999. The "odds" of having a particular exchange is P = 1/512.
The number of four digit phone numbers, "defg" is exactly the same except that 'd', 'e', 'f', 'g' can
EACH INDEPENDENTLY have one of the ten integer values: 0, 1, 2, ..., 9. The total number of possible
4digit phone numbers "defg", starting with: 0000, 0001, 0002, ..., 6825, ..., 9999. So there are 10
x 10 x 10 x 10 = 10^4 four digit integers between "0000" and "9999" with no other restrictions on the
counting procedure. Combining that with the 512 possible exchanges gives: 5,120,000 possible phone numbers,
each with a probability: P = 1/5,120,000.
Permutations and/or combinations do not enter into the counting procedure here because there is no
"identifier" that excludes some subset of the telephone numbers from the counting process. The verbal
statement of this nonexclusion in the inquiry is: "...neither phone number is specified in advance."
(except that neither '0' nor '1' can appear in any of the 3digit exchanges).
A similar example is, "How many license plates can be made using two letters followed by a 3digit number:
LMabcd, with no other restrictions?"
For 'L' and 'M' there are 26 possible choices each, and for 'a', 'b', 'c', 'd' there are 10 choices each
(digits 0, 1, 2, ..., 9). So there are: 26 x 26 x 10 x 10 x 10 = 676,000 possible license plates.
Vince Calder
Click here to return to the Mathematics Archives
 
Update: June 2012

