GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Sep 2019, 21:36

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 27 May 2010
Posts: 200
How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,  [#permalink]

### Show Tags

25 Jun 2019, 23:04
8
00:00

Difficulty:

45% (medium)

Question Stats:

71% (03:08) correct 29% (02:29) wrong based on 41 sessions

### HideShow timer Statistics

How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 such that odd numbers occur at even places.

(A) 160
(B) 175
(C) 180
(D) 220
(E) 200

_________________
Please give Kudos if you like the post
##### Most Helpful Community Reply
Senior Manager
Joined: 16 Jan 2019
Posts: 444
Location: India
Concentration: General Management
WE: Sales (Other)
Re: How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,  [#permalink]

### Show Tags

25 Jun 2019, 23:42
6
1
There are 4 even places out of which we must choose 3 for the odd numbers = 4C3

In these 3 chosen places, the odd numbers can be arranged in 3!/2! ways

Now the rest of the numbers can be arranged in 6!/4!2! ways

Total number of ways = 4C3*3!/2!*6!/4!2! = 180

Answer is (C)

nick1816 can you confirm if this solution is right?
##### General Discussion
Director
Joined: 19 Oct 2018
Posts: 853
Location: India
Re: How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,  [#permalink]

### Show Tags

25 Jun 2019, 23:20
1
Case 1- when odd numbers and 4 occurs at even place.
Number of 9 digits can be formed= $$4C3*\frac{3!}{2!}*\frac{5!}{4!}$$= 4*3*5=60

Case 2- when odd numbers and 2 occurs at even place.
Number of 9 digits can be formed= $$4C3*\frac{3!}{2!}*\frac{5!}{3!2!}$$= 4*3*10=120

Total= 60+120=180

prashanths wrote:
How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 such that odd numbers occur at even places.

(A) 160
(B) 175
(C) 180
(D) 220
(E) 200
Director
Joined: 19 Oct 2018
Posts: 853
Location: India
Re: How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,  [#permalink]

### Show Tags

25 Jun 2019, 23:50
1
Bhai 1 number!!!

firas92 wrote:
There are 4 even places out of which we must choose 3 for the odd numbers = 4C3

In these 3 chosen places, the odd numbers can be arranged in 3!/2! ways

Now the rest of the numbers can be arranged in 6!/4!2! ways

Total number of ways = 4C3*3!/2!*6!/4!2! = 180

Answer is (C)

nick1816 can you confirm if this solution is right?
Senior Manager
Joined: 15 Feb 2018
Posts: 343
Re: How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,  [#permalink]

### Show Tags

18 Aug 2019, 13:52
1
I spent too long trying to figure out the combinations for the even placements. I instead looked to the odd placements

6 even numbers can go in position one
5 in three
4 in five
3 in seven
2 in nine
This is 6!

2 is repeated four times and 4 repeated twice, so we need to reduce the factorial

6!/2!4!=15. The answer choices must be divisible by 15. Only C is.

C
Intern
Joined: 20 Jun 2019
Posts: 40
Re: How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,  [#permalink]

### Show Tags

19 Aug 2019, 22:18
1
There are 4 odd places and 3 odd numbers.
1,1,3,X where X is the empty even place which will not have the odd number.
Number of ways to arrange these = 4!/2 = 12

After the 3 odd numbers, there are 6 even numbers remaining for the 6 places.
Number of ways to arrange these = 6!/ (4!*2!) = 15

Hence total number of ways to arrange = 15 * 12 = 180

C
Re: How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,   [#permalink] 19 Aug 2019, 22:18
Display posts from previous: Sort by

# How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2,

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne