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?

- 1809
- 7569
- 9369
- 8649
- 8289

@ INR

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.

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.,

**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 9C_{2} 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., 9C_{2} × 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 9C_{2} = \\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 9C_{3} = \\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**

Try it free!

Register in 2 easy steps and

Start learning in 5 minutes!

21. GMAT 700 Level Sample Question in Counting Methods

22. GMAT Hard Math Arithmetic | Permutation Combination - Selections | GMAT Problem Solving

23. GMAT 700 800 Level Question | Arithmetic | Counting Methods Problem Solving

24. GMAT Hard Math | Arithmetic | Number Properties Sample Question

25. Geometry | Coordinate Geometry DS Practice | GMAT Challenging Math Question

26. GMAT Data Sufficiency | Challenging Number Properties & Inequalities Question

27. GMAT DS Sample Question | Statistics | Mean Median Range

28. GMAT Challenging Question | Arithmetic | Numbers & Progressions DS

29. GMAT 650 Level Algebra Question | Polynomials

30. GMAT Hard Math Questions | Arithmetic | Number Properties DS

Copyrights © 2016 - 21 All Rights Reserved by Wizako.com - An Ascent Education Initiative.

Privacy Policy | Terms & Conditions

GMAT^{®} is a registered trademark of the Graduate Management Admission Council (GMAC). This website is not endorsed or approved by GMAC.

GRE^{®} is a registered trademarks of Educational Testing Service (ETS). This website is not endorsed or approved by ETS.

SAT^{®} is a registered trademark of the College Board, which was not involved in the production of, and does not endorse this product.

Wizako - GMAT, GRE, SAT Prep

An Ascent Education Initiative

48/1 Ramagiri Nagar

Velachery Taramani Link Road.,

Velachery, Chennai 600 042. India

**Phone:** (91) 44 4500 8484

**Mobile:** (91) 95000 48484

**WhatsApp:** WhatsApp Now

**Email:** learn@wizako.com

Leave A Message