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

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

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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

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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

_________________
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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]

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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

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]

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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]

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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

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]

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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]

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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
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