Problem Solving

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, would you please explain the portion in red above?

My reasoning:
Say if I take any number P from 81 two digit numbers with distinct integers then to choose Q i have only 80 ways. Also, if I choose 53 then another number with both digits in common i.e, is 35 but still these are 2 choices!
Can you please explain?

winniethepooh 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, would you please explain the portion in red above?

My reasoning:
Say if I take any number P from 81 two digit numbers with distinct integers then to choose Q i have only 80 ways. Also, if I choose 53 then another number with both digits in common i.e, is 35 but still these are 2 choices!
Can you please explain?

In the red portion, we are counting ALL of the ways to choose integers P and Q from set S.
There are NO restrictions here.
P and Q can have digits in common.
P and Q can be the same integer.
NO restrictions.
Thus, the number of choices for P is 81, and the number of choices for Q also is 81.
Number of ways to choose P and Q = 81*81 = 6561.

From this result, we then SUBTRACT the number of BAD combinations in which P and Q have NO digits in common: 4032.

Thus, the number of ways to choose P and Q so that they have at least one digit in common = 6561-4032 = 2529.

GMATGuruNY wrote:

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.

Hey Mitch.though I agree with you,but still have difficulty in picturing clearly the logic you explained above in blue.TIA

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.

Gosh dang it.

I got 2592, which is close to 2529, and it literally took me like 8 minutes to get that. I would've picked "D" on the real GMAT, but I'm rather pissed that this question almost completely stumped me.

This is what I did:

Tens = 1 to 9 (9 choices)
Units = 0 to 9 - digit used (10-1 = 9 choices)

Total = 9 x 9 = 81 possible combinations of P.

Total combos = 81 x 81

Answer must = at least 1 digit in common. 1 digit in common = 1 in common or 2 in common, which means answer = (total combinations - digits with 0 digits in common)

Digits with 0 in common = 81 x 7 (any digit from 1 to 9 minus 2 already used) x 7 (any digit from 0 to 9 minus the 3 already used).

(81 x 81) - (81 x 7 x 7) = 2592

Which is almost 2529, so that's the answer I would've picked. But I must've done something wrong.

math wizards advise please?

Digits with 0 in common = 81 x 7 (any digit from 1 to 9 minus 2 already used) x 7 (any digit from 0 to 9 minus the 3 already used).
plz explain this

can problems like these solved within a min ??

in my opinion if I'm not clear or close to answer by 1 min, then i shud make an educated guess and move on.

Its was very tough..

i tried everything i know and got the answer near to D and took 10 mins..

Best thing for such questions guess and go

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.

Can you please explain why the order of choosing the combinations is Tens digit first for P and Q and then the units digit. If i choose tens and units of P alone i get 9 * 9 ways of getting P. While Q will now have 7 * 7. That doesnt match your answer. Kindly explain?

ssidda01 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 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.

Can you please explain why the order of choosing the combinations is Tens digit first for P and Q and then the units digit. If i choose tens and units of P alone i get 9 * 9 ways of getting P. While Q will now have 7 * 7. That doesnt match your answer. Kindly explain?

Start with the MOST RESTRICTED positions.
Neither TENS DIGIT can be 0; the units digits are NOT subject to this restriction.
It is for this reason that I counted FIRST the number of options for each TENS DIGIT.

If we first count the number of options for either units digit, we must recognize a key issue:
Selecting a UNITS DIGIT OF 0 means that we DON'T lose an option for the next tens digit we select, since a tens digit cannot be 0.
Selecting a NONZERO UNITS DIGIT means that we DO lose one option for the next tens digit we select.
Thus, your approach requires that we consider TWO cases:

Case 1: The units digit of P is 0
Units digit of P is 0 = 1 choice.
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 P's tens digit = 8 choices.
Units digit of Q can be any digit 1 through 9, excluding P's tens digit and Q's tens digit = 7 choices.
Multiplying our choices for each digit, we get:
1*9*8*7 = 504.

Case 2: The units digit of P is not 0
Units digit of P can be any digit 1 through 9 = 9 choices.
Tens digit of P can be any digit 1 through 9, excluding the digit already used = 8 choices.
Tens digit of Q can be any digit 1 through 9, excluding the 2 digits already used = 7 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:
9*8*7*7 = 3528.

Total options = 504+3528 = 4032.

E in answer.

A quick way to solve is as follows.

As the answer involves ordered pairs it must be divisible by 2.

Further the number 9 must be a factor of the answer as there are 72 numbers with non zero digits and 9 numbers with one of the digits as a zero.

Here we can have pairs like (12,12),(13,13),...,(98,98)

so we need to add 81 to 2448

and the answer is 2529.

I sense That i will not see a question of this difficulty on the test haha.

IMO D took me 4 mins!

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 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.

Hello !!!
I have a small doubt. There are 91 total two digit numbers. Of these, we don't want 11,22,33,44,55,66,77,88,99. So we are left with (91-9) = 82 two digit distinct digit numbers. So set S should have 82 numbers right ?

prateek9567 wrote:
Hello !!!
I have a small doubt. There are 91 total two digit numbers. Of these, we don't want 11,22,33,44,55,66,77,88,99. So we are left with (91-9) = 82 two digit distinct digit numbers. So set S should have 82 numbers right ?

Just to help you with a small observation

AMONG POSITIVE INTEGERS


Single digit Numbers = 9 [{1 through 9}]
Two digit Numbers = 90 [{1 through 99} - {1 through 9} = 99-9 = 90]
Three digit Numbers = 900 [{1 through 999} - {1 through 99} = 999-99 = 900]
Four digit Numbers = 9000 [{1 through 9999} - {1 through 999} = 9999-999 = 9000]

and so on...