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# David and Rachel are getting married. The extended family wants to hav

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Retired Moderator
Joined: 29 Apr 2015
Posts: 798
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Concentration: Economics, Finance
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David and Rachel are getting married. The extended family wants to hav  [#permalink]

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02 Jul 2015, 11:04
2
7
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Difficulty:

55% (hard)

Question Stats:

62% (01:46) correct 38% (01:46) wrong based on 124 sessions

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David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?

A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!
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Joined: 02 Sep 2009
Posts: 62287
Re: David and Rachel are getting married. The extended family wants to hav  [#permalink]

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02 Jul 2015, 11:14
reto wrote:
David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?

A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!

In how many ways David's father and Rachel's mother can stand together? Consider them as one unit: {F, M}. So, we'd have 9 units {F, M}, 1, 2, 3, 4, 5, 6, 7, 8, which can be arranged in 9! ways. David's father and Rachel's mother within their unit can be arranged in 2! ways. Therefore, David's father and Rachel's mother can stand together in 9!*2! ways.

Total - Restriction =

= 10! - 9!*2! =

= 9!(10 - 2!) =

= 8*9!.

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Re: David and Rachel are getting married. The extended family wants to hav  [#permalink]

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20 Jul 2015, 11:58
2
reto wrote:
David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?

A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!

I just wanted to add how you get to the final 8x9! because I was struggling with it:

Total Options are 10!
Forbidden Options are 9!*2

Good Options are 10!-9!*2 which can be rewritten as:

10*9! - 2*9!

It is then possible to extract 9! as the common factor:

9!(10-2) = 9! * 8
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Re: David and Rachel are getting married. The extended family wants to hav  [#permalink]

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20 Jul 2015, 12:07
1
reto wrote:
reto wrote:
David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?

A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!

I just wanted to add how you get to the final 8x9! because I was struggling with it:

Total Options are 10!
Forbidden Options are 9!*2

Good Options are 10!-9!*2 which can be rewritten as:

10*9! - 2*9!

It is then possible to extract 9! as the common factor:

9!(10-2) = 9! * 8

Additionally, I like using the slot method to look at the possible, allowed and forbidden cases.

Consider 10 slots for 10 members.

_ _ _ _ _ _ _ _ _ _ We can arrange these 10 members in 10! ways (=10P10).

NOw consider 1 scenario in which the dad (D) stands next to mom (M) : DM + 8 slots for the rest of the members: D M _ _ _ _ _ _ _ _ , we can arrange these in 9! X 2! ways ( 2! ways for arranging D and M into DM or MD and 9! to arrange 8! members + 1 combined (D and M) slots)

Thus we have the total number of arrangements possible = 10! - 9!X2 = 10X9! - 9!X2 = 9! (8), C is the correct answer.
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Re: David and Rachel are getting married. The extended family wants to hav  [#permalink]

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21 Mar 2017, 02:03
reto wrote:
David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?

A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!

10 member can be arranged in 10! ways
Let those two stand together. We will consider them as one entity with 2! way to arrange within themselves

total number of people = 8 +1 = 9. ways of arranging - 9!

therefore total ways of arranging with this restriction = 10! - 9!2! = 8.9!
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Re: David and Rachel are getting married. The extended family wants to hav  [#permalink]

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08 Feb 2020, 11:48
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Re: David and Rachel are getting married. The extended family wants to hav   [#permalink] 08 Feb 2020, 11:48
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