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by goyalsau » Tue Dec 07, 2010 9:30 am
Let S be the set of all the two-digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P and Q belong to S and have at least one digit in common?



4032

2720

2439

2529

2448
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by Rahul@gurome » Tue Dec 07, 2010 12:01 pm
Number of two-digit natural numbers with distinct digits = 9*9 = 81. Thus S contains 81 natural numbers.

Now let's see for a particular P how many Q's are there.

Say P = 53. Then Q may be any natural number in S which contains 3 or 5 or both. Number of such natural numbers = (Number of natural numbers with distinct digits starting with 3 + Number of natural numbers with distinct digits starting with 5 + Number of natural numbers with distinct digits ending with 3 + Number of natural numbers with distinct digits ending with 5)

1. Number of natural numbers with distinct digits starting with 3 = 9 (33 is not counted)
2. Number of natural numbers with distinct digits starting with 5 = 9 (55 is not counted)
3. Number of natural numbers with distinct digits ending with 3 = 7 (33 is not counted and 53 already counted in 2)
4. Number of natural numbers with distinct digits ending with 5 = 7 (55 is not counted and 35 already counted in 1)

Total number of possible Q = (9 + 9 + 7 + 7) = 32

But if P = 10, 20, 30 etc (P containing zero), then the calculation will be different.

Say P = 20. Then Q may be any natural number in S which contains 2 or 0 or both. Number of such natural numbers = (Number of natural numbers with distinct digits starting with 2 + Number of natural numbers with distinct digits ending with 0 + Number of natural numbers with distinct digits ending with 2)

1. Number of natural numbers with distinct digits starting with 2 = 9 (22 is not counted)
2. Number of natural numbers with distinct digits ending with 0 = 8 (20 already counted in 1)
3. Number of natural numbers with distinct digits ending with 2 = 8 (22 is not counted)

Total number of possible Q = (9 + 8 + 8) = 25

Total number of P such that P contains 0 = 9 => Number of possible ordered pair (P, Q) = 9*25 = 225
Total number of P such that P does not contains 0 = (81 - 9) = 72 => Number of possible ordered pair (P, Q) = 72*32 = 2304

Total number of possible ordered pair (P, Q) = (2304 + 225) = 2529

The correct answer is D.
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by diebeatsthegmat » Wed Dec 08, 2010 11:21 am
goyalsau wrote:Let S be the set of all the two-digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P and Q belong to S and have at least one digit in common?



4032

2720

2439

2529

2448
hi, may i know what source this PS from?
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by goyalsau » Wed Dec 08, 2010 9:21 pm
diebeatsthegmat wrote: hi, may i know what source this PS from?
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by gmatusa2010 » Sat Dec 11, 2010 1:34 am
what is an ordered pair? this is the first time I've heard this tested in gmat thats not in the context of coordinate geometry.
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by shovan85 » Sat Dec 11, 2010 1:39 am
gmatusa2010 wrote:what is an ordered pair? this is the first time I've heard this tested in gmat thats not in the context of coordinate geometry.
In mathematics, an ordered pair is a collection (of objects) having two entries (or or coordinates ), such that it is distinguishable, which object is the first entry (or first coordinate ) of the pair and which object is the second entry (or second coordinate ) of the pair.

If the first coordinate is a and the second is b, the usual notation for an ordered pair is (a, b). The pair is "ordered" in that (a, b) differs from (b, a) unless a = b.
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by prachich1987 » Sat Dec 11, 2010 2:55 am
Rahul@gurome wrote:Number of two-digit natural numbers with distinct digits = 9*9 = 81. Thus S contains 81 natural numbers.

Now let's see for a particular P how many Q's are there.

Say P = 53. Then Q may be any natural number in S which contains 3 or 5 or both. Number of such natural numbers = (Number of natural numbers with distinct digits starting with 3 + Number of natural numbers with distinct digits starting with 5 + Number of natural numbers with distinct digits ending with 3 + Number of natural numbers with distinct digits ending with 5)

1. Number of natural numbers with distinct digits starting with 3 = 9 (33 is not counted)
2. Number of natural numbers with distinct digits starting with 5 = 9 (55 is not counted)
3. Number of natural numbers with distinct digits ending with 3 = 7 (33 is not counted and 53 already counted in 2)
4. Number of natural numbers with distinct digits ending with 5 = 7 (55 is not counted and 35 already counted in 1)

Total number of possible Q = (9 + 9 + 7 + 7) = 32

But if P = 10, 20, 30 etc (P containing zero), then the calculation will be different.

Say P = 20. Then Q may be any natural number in S which contains 2 or 0 or both. Number of such natural numbers = (Number of natural numbers with distinct digits starting with 2 + Number of natural numbers with distinct digits ending with 0 + Number of natural numbers with distinct digits ending with 2)

1. Number of natural numbers with distinct digits starting with 2 = 9 (22 is not counted)
2. Number of natural numbers with distinct digits ending with 0 = 8 (20 already counted in 1)
3. Number of natural numbers with distinct digits ending with 2 = 8 (22 is not counted)

Total number of possible Q = (9 + 8 + 8) = 25

Total number of P such that P contains 0 = 9 => Number of possible ordered pair (P, Q) = 9*25 = 225
Total number of P such that P does not contains 0 = (81 - 9) = 72 => Number of possible ordered pair (P, Q) = 72*32 = 2304

Total number of possible ordered pair (P, Q) = (2304 + 225) = 2529

The correct answer is D.


Thanks for the above explanation.
But I don't think such lengthy problems appear in GMAT.
Plz advise
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by thebigkats » Sat Dec 25, 2010 8:07 pm
Hi:
per definition there is a set of no's (xy) wherein x=1..9 and y=0..9 and x!=y
for x=1..9 and y=0..9 ==> xy=10..99 ==> total of 90 numbers
in this set remove no's where x=y (e.g. 11, 22, 33,44,55,66,77,88,99)
So the set S = 90-9 = 81 no's that meet condition x=1..9 and y=0..9 and x!=y

Now for every xy, we need to find pq wherein "pq=(x, 0..9) or (1..9, x) or (y, 0..9) or (1..9, y). Lets take each pq possibility

pq = (x, 0..9) =>10 combinations ==> but we can't take the one where q==x. so we need to cancel that one
hence 9 combinations
pq = (1..9, x) ==> 9 combos ==> again we can't take the one where p ==x. so we need to cancel that one
hence 8 combinations ==> also, one of these combos would have been counted above already. so we need to cancel that one also
hence 7 combos
pq = (y, 0..9) ==> 10 combinations ==> but we can't take the one where q==x. so we need to cancel that one
hence 9 combinations (EXCEPTION - when y = 0, total == 0 combos because 0 can't be at beginning)
pq = (1..9, y) ==> 9 combos ==> again we can't take the one where p ==x. so we need to cancel that one
hence 8 combinations ==> also, one of these combos would have been counted above already. so we need to cancel that one also
hence 7 combos

So for a given (xy) total combinations = 9 + 7 + 9 + 7 = 32 OR 25 combos depending on whether y = 0 OR 1..9

Out of total 81 (xy) there are 9 numbers wherein y==0
So for those 9 numbers, combos = 25
For other 72 xy, combos = 32

So total = 32*72 + 9 *25
==> 2529
Last edited by thebigkats on Thu May 05, 2011 8:20 pm, edited 1 time in total.
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by needthis » Thu May 05, 2011 9:23 am
This is a tricky one!
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by GMATGuruNY » Thu May 05, 2011 7:33 pm
goyalsau wrote:Let S be the set of all the two-digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P and Q belong to S and have at least one digit in common?



4032

2720

2439

2529

2448
Number of integers in S:
Tens digit can be any digit 1 through 9 = 9 choices.
Units digit can be any digit 0 though 9, excluding the digit used for the tens digit = 9 choices.
Multiplying our choices for each digit, we get:
Total integers in S = 9*9 = 81.

Combinations of P and Q that have at least 1 digit in common = Total ways to combine P and Q - Combinations of P and Q that have no digits in common.

Total ways to combine P and Q:
Number of choices for P = 81.
Number of choices for Q = 81.
Total combinations = 81*81 = 6561.

Combinations of P and Q that have no digits in common:
Tens digit of P can be any digit 1 through 9 = 9 choices.
Tens digit of Q can be any digit 1 through 9, excluding the digit used for the tens digit of P = 8 choices.
Units digit of P can be any digit 0 through 9, excluding the 2 digits already used = 8 choices.
Units digit of Q can be any digit 0 through 9, excluding the 3 digits already used = 7 choices.
Multiplying our choices for each digit, we get:
Combinations with no digits in common = 9*8*8*7 = 4032.

Thus, combinations of P and Q that have at least 1 digit in common = 6561-4032 = 2529.

The correct answer is D.
Last edited by GMATGuruNY on Fri Sep 21, 2012 9:44 am, edited 1 time in total.
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by finance » Sat May 07, 2011 9:03 am
GMATGuruNY wrote:
goyalsau wrote:Let S be the set of all the two-digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P and Q belong to S and have at least one digit in common?



4032

2720

2439

2529

2448
Number of integers in S:
Tens digit can be any digit 1 through 9 = 9 choices.
Units digit can be any digit 0 though 9, excluding the digit used for the tens digit = 9 choices.
Multiplying our choices for each digit, we get:
Total integers in S = 9*9 = 81.

Combinations of P and Q that have at least 1 digit in common = Total ways to combine P and Q - Combinations of P and Q that have no digits in common.

Total ways to combine P and Q:
Number of choices for P = 81.
Number of choices for Q = 81.
Total combinations = 81*81 = 6561.

Combinations of P and Q that have no digits in common:
Tens digit of P can be any digit 1 through 9 = 9 choices.
Tens digit of Q can be any digit 1 through 9, excluding the digit used for the tens digit of P = 8 choices.
Units of digit of P can be any digit 0 through 9, excluding the 2 digits already used = 8 choices.
Units of digit of Q can be any digit 0 through 9, excluding the 3 digits already used = 7 choices.
Multiplying our choices for each digit, we get:
Combinations with no digits in common = 9*8*8*7 = 4032.

Thus, combinations of P and Q that have at least 1 digit in common = 6561-4032 = 2529.

The correct answer is D.
Great explanation! Thank you!
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by prashant misra » Mon Oct 17, 2011 6:56 am
thank you gmatguruNY i was not able to understand the question how others have solved but it was much easier to understand the way you solved the question
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by anuu » Mon Oct 17, 2011 8:45 am
GMATGuruNY wrote:
goyalsau wrote:Let S be the set of all the two-digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P and Q belong to S and have at least one digit in common?



4032

2720

2439

2529

2448
Number of integers in S:
Tens digit can be any digit 1 through 9 = 9 choices.
Units digit can be any digit 0 though 9, excluding the digit used for the tens digit = 9 choices.
Multiplying our choices for each digit, we get:
Total integers in S = 9*9 = 81.

Combinations of P and Q that have at least 1 digit in common = Total ways to combine P and Q - Combinations of P and Q that have no digits in common.

Total ways to combine P and Q:
Number of choices for P = 81.
Number of choices for Q = 81.
Total combinations = 81*81 = 6561.

Combinations of P and Q that have no digits in common:
Tens digit of P can be any digit 1 through 9 = 9 choices.
Tens digit of Q can be any digit 1 through 9, excluding the digit used for the tens digit of P = 8 choices.
Units of digit of P can be any digit 0 through 9, excluding the 2 digits already used = 8 choices.
Units of digit of Q can be any digit 0 through 9, excluding the 3 digits already used = 7 choices.
Multiplying our choices for each digit, we get:
Combinations with no digits in common = 9*8*8*7 = 4032.

Thus, combinations of P and Q that have at least 1 digit in common = 6561-4032 = 2529.

The correct answer is D.


Hi Mitch,

I've doubt in the folowing section :'Combination of P and Q that have no digits in common':

Can we consider the following case :


tens digits of P = 9 choices
unit digit of P = 9 choices(excluding the digit used for the tens digit)


Similarly for Q

Tens digit of Q = 7 choices (total 9 choices - 2 (the units and tens digits used for p))
Unit digit of Q = 7 choices (total 10 choices - 3 (the units and tens digits used for p &the tens))

Combinations with no digits in common = 9*9*7*7= 3969

Total combinations of P and Q that have at least 1 digit in common = 6561-3969 = 2592


Thank you,
Anu
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by GMATGuruNY » Mon Oct 17, 2011 9:02 am
anuu wrote:
GMATGuruNY wrote:
goyalsau wrote:Let S be the set of all the two-digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P and Q belong to S and have at least one digit in common?



4032

2720

2439

2529

2448
Number of integers in S:
Tens digit can be any digit 1 through 9 = 9 choices.
Units digit can be any digit 0 though 9, excluding the digit used for the tens digit = 9 choices.
Multiplying our choices for each digit, we get:
Total integers in S = 9*9 = 81.

Combinations of P and Q that have at least 1 digit in common = Total ways to combine P and Q - Combinations of P and Q that have no digits in common.

Total ways to combine P and Q:
Number of choices for P = 81.
Number of choices for Q = 81.
Total combinations = 81*81 = 6561.

Combinations of P and Q that have no digits in common:
Tens digit of P can be any digit 1 through 9 = 9 choices.
Tens digit of Q can be any digit 1 through 9, excluding the digit used for the tens digit of P = 8 choices.
Units of digit of P can be any digit 0 through 9, excluding the 2 digits already used = 8 choices.
Units of digit of Q can be any digit 0 through 9, excluding the 3 digits already used = 7 choices.
Multiplying our choices for each digit, we get:
Combinations with no digits in common = 9*8*8*7 = 4032.

Thus, combinations of P and Q that have at least 1 digit in common = 6561-4032 = 2529.

The correct answer is D.


Hi Mitch,

I've doubt in the folowing section :'Combination of P and Q that have no digits in common':

Can we consider the following case :


tens digits of P = 9 choices
unit digit of P = 9 choices(excluding the digit used for the tens digit)


Similarly for Q

Tens digit of Q = 7 choices (total 9 choices - 2 (the units and tens digits used for p))
Unit digit of Q = 7 choices (total 10 choices - 3 (the units and tens digits used for p &the tens))
The portion in red overlooks the following:
If the units digit of P = 0, then the number of options for the tens digit of Q = 8 (because we can use any digit other than the two digits used in P).

My solution bypasses this issue by FIRST counting the number of options for each hundreds digit -- neither of which can be 0 -- and THEN counting the number of options for each units digit.
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by ArunangsuSahu » Sun Jan 01, 2012 9:55 am
Total 2-digit numbers = 99-10+1=90

11,22,33,.....99 are to be discarded . SO total 90-9=81

Combinations of P and Q that have at least 1 digit in common = Total ways to combine P and Q - Combinations of P and Q that have no digits in common.

No Digits in Common pattern:

Let's take 10-19 (excluding 11)
from 20-29---for 10 (10,21 and obviously as per question are to be excluded)

So, for one set Combinations = 9C1*7C1*8 [Total 8 series can partner with 10-19]

so for 9 set of Combinations = 9*9C1*7C1*8=4032

Without Restrictions = 81C1*81C1=6561

Therefore at least one digit in Common = 6562-4032=2529

(D) is the answer
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