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The number ways in which n distinct objects can be put into

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The number ways in which n distinct objects can be put into [#permalink] New post 26 Jun 2013, 02:16
The number ways in which n distinct objects can be put into two distinct boxes so that no box remains empty, is

(A) 2^(n)-1
(B) n^2 -1
(C) 2^n -2
(D) n^2 -2
(E) n!


Please Explain!
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Re: The number ways in which n distinct objects can be put into [#permalink] New post 26 Jun 2013, 02:25
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The number ways in which n distinct objects can be put into two distinct boxes so that no box remains empty, is
(A) 2^(n)-1
(B) n^2 -1
(C) 2^n -2
(D) n^2 -2
(E) n!

1st object can b place in 2 ways (either of the boxes)
Similarly, 2nd object can b place in 2 ways
Similarly, 3rd object can b place in 2 ways
Similarly, 4th object can b place in 2 ways
.
.
.
.Similarly, nth object can b place in 2 ways

Because all placement of 2nd object is independent of placement of 1st object, the total no of ways in n objects can be place in two boxes = 2 x 2 x 2 x 2.............2(nth)
= \(2^n\)

But this total no ways also include the possibility - 2 cases
1) that all the objects are placed in only box1
2) that all the objects are placed in only box2

Because we want all the boxed to be filled we need to reduce the total no of ways by 2. So
\(2^n\) - 2

Hope this helps many
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The probabilty that a leap year selected [#permalink] New post 26 Jun 2013, 03:22
The probabilty that a leap year selected at random contains either 53 Sundays or 53 Mondays, is

(A) 2/7
(B) 4/7
(C) 3/7
(D) 1/7
(E) 5/7

Please expalin
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Re: The probabilty that a leap year selected [#permalink] New post 26 Jun 2013, 04:30
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The probabilty that a leap year selected at random contains either 53 Sundays or 53 Mondays, is
(A) 2/7
(B) 4/7
(C) 3/7
(D) 1/7
(E) 5/7

Very good question.

Leap year only has 1 extra day over the normal year in month of FEB.
A leap year has 52 weeks of 7 days (which means 52 Sundays & 52 Mondays) plus two more days.

The question basically asks the composition of the remaining two days
The Remaining two days can be
Mon Tue (Our case- Contains Extra Monday)
Tue Wed
Wed Thur
Thur Fri
Fri Sat
Sat Sun (Our case- Contains Extra Sunday)
Sun Mon (Our case- Contains Extra Sunday & Monday)

So the probability is = 3/7

Hope this helps.
Fame
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Re: The probabilty that a leap year selected [#permalink] New post 26 Jun 2013, 05:57
fameatop wrote:
The probabilty that a leap year selected at random contains either 53 Sundays or 53 Mondays, is
(A) 2/7
(B) 4/7
(C) 3/7
(D) 1/7
(E) 5/7

Very good question.

Leap year only has 1 extra day over the normal year in month of FEB.
A leap year has 52 weeks of 7 days (which means 52 Sundays & 52 Mondays) plus two more days.

The question basically asks the composition of the remaining two days
The Remaining two days can be
Mon Tue (Our case- Contains Extra Monday)
Tue Wed
Wed Thur
Thur Fri
Fri Sat
Sat Sun (Our case- Contains Extra Sunday)
Sun Mon (Our case- Contains Extra Sunday & Monday)

So the probability is = 3/7

Hope this helps.
Fame



Hey why cant it be 4/7.?
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Re: The probabilty that a leap year selected [#permalink] New post 27 Jun 2013, 20:20
Expert's post
akijuneja wrote:
fameatop wrote:
The probabilty that a leap year selected at random contains either 53 Sundays or 53 Mondays, is
(A) 2/7
(B) 4/7
(C) 3/7
(D) 1/7
(E) 5/7

Very good question.

Leap year only has 1 extra day over the normal year in month of FEB.
A leap year has 52 weeks of 7 days (which means 52 Sundays & 52 Mondays) plus two more days.

The question basically asks the composition of the remaining two days
The Remaining two days can be
Mon Tue (Our case- Contains Extra Monday)
Tue Wed
Wed Thur
Thur Fri
Fri Sat
Sat Sun (Our case- Contains Extra Sunday)
Sun Mon (Our case- Contains Extra Sunday & Monday)

So the probability is = 3/7

Hope this helps.
Fame



Hey why cant it be 4/7.?


Note that a leap year has 366 days i.e. 52 complete weeks + 2 extra days.
We need either a Sunday or a Monday or (both) on the two extra days- 365th and 366th days. Think about it this way:

The 365th day can be any of the 7 days. We need it to be Saturday/Sunday/Monday. In these 3 cases, we will have Sunday or Monday or both on the two extra days. If the 365th day is any of the other 4 days, we will not have a Sunday or a Monday.
So the required probability is 3/7.
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Re: The probabilty that a leap year selected   [#permalink] 27 Jun 2013, 20:20
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