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 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.

Really the most lucid and convenient solution among all !! Thanx a ton !!

The correct answer is D.

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.

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.