# GMAT 700 800 Level Quant Question 23

###### Counting Methods & Numbers | Permutation Combination Practice | GMAT Sample Questions

The given question is a challenging GMAT 700 800 level quant problem solving question combining concepts in permutation combination (counting methods) and number properties. Another interesting practice question to understand how to list down possibilities and then count outcomes for each of those possibilities.

Question 23: How many 6-digit numbers can be formed using the digits {1, 2, 3, ... 9} such that any digit that appears in such a number appears at least twice?

1. 1809
2. 7569
3. 9369
4. 8649
5. 8289

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### Explanatory Answer | GMAT Permutation Combination Practice Question

#### Key Data

1. 6-digit numbers.
2. Formed using digits {1, 2, 3,..., 9}. Note: Does not include zero.
3. Any digit that appears should appear at least twice.

#### Examples: Those that satisfy and those that do not

Some 6-digit numbers that satisfy the condition: 555555, 223344, 111999, etc.,
Some 6-digit numbers that do not satisfy the condition: 123456, 123444, 558812, etc.,

#### List Down Possibilities and Count

Possibility 1: All 6 digits are same
Example: 111111
9 such numbers possible.

Possibility 2: 4 digits show one value and 2 digits show another value. Example: 373777
Step 1: We are selecting 2 digits from 9 numbers. This can be done in 9C2 ways.
Step 2: For example, if the digits are 3 and 7, either 3 appears 4 times and 7 appears twice or vice versa.
So, there are 2 possibilities.
Step 3: Reordering of 6 digits can be done in $$frac{6!}{4! × 2!}$ = 15 Number of such numbers = Product of values obtained in the above 3 steps. i.e., 9C2 × 2 × 15 = $\frac{9 × 8}{1 × 2}$ × 2 × 15 = 1080 Possibility 3: 3 digits show one value and another 3 digits show a second value. Example: 444777 Step 1: We are selecting 2 digits from 9 numbers. This can be done in 9C2 = $\frac{9 × 8}{1 × 2}$ = 36 ways. Step 2: Reordering of 6 digits can be done in $\frac{6!}{3! × 3!}$ = $\frac{6 × 5 × 4 × 3!}{3! × 3!}$ = 20 ways Number of such numbers = 36 × 20 = 720 Possibility 4: 3 different digits, each appearing twice. Example: 234234 Step 1: We are selecting 3 digits from 9 numbers. This can be done in 9C3 = $\frac{9 × 8 × 7}{1 × 2 × 3}$ = 84 ways. Step 2: Reordering of 6 digits can be done in $\frac{6!}{2! × 2! × 2!}$ = $\frac{720}{8}$ = 90 ways Number of such numbers = 84 × 90 = 7560 Total such numbers =$9 + 1080 + 720 + 7560) = 9369

##### Choice C is the correct answer.

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