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danlew
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Sam and Jane go to a birthday party of a friend. They have an argument 15 Jun 2017, 03:10
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C
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63% (01:46) correct 37% (01:42) wrong based on 230 sessions

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Sam and Jane meet each other at a party. They have an argument in the party and do not talk to each other for the rest of the party. A photographer wants to take a picture of all the guests at the party. How many options does the photographer have to arrange the 10 guests at the party, if Jane does not want to stand next to Sam when the picture is taken?

A. 9!
B. 8!x9
C. 8x9!
D. \(\frac{10!}{2!}\)
E. 10!
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C
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Sam and Jane go to a birthday party of a friend. They have an argument 15 Jun 2017, 03:16
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danlew
Sam and Jane meet each other at a party. They have an argument in the party and do not talk to each other for the rest of the party. A photographer wants to take a picture of all the guests at the party. How many options does the photographer have to arrange the 10 guests at the party, if Jane does not want to stand next to Sam when the picture is taken?

A. 9!
B. 8!x9
C. 8x9!
D. \(\frac{10!}{2!}\)
E. 10!
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Glue those two together. We'd have 9 units: 1, 2, 3, 4, 5, 6, 7, 8, and {Sam, Jane}. These 9 units can be arranged in 9! ways. Sam and Jane within their unit can be arranged in 2! ways. Thus the total number of ways to arrange 10 people so that Sam and Jane to stand together is 9!*2!.

10 people without any restriction can be arranged in 10! ways, thus in 10! - 9!*2! arrangements Sam and Jane will not be standing together.

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

Answer: C.
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Sam and Jane go to a birthday party of a friend. They have an argument 15 Jun 2017, 03:17
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Total ways of arranging is 10! And the way when sam and jane are together is 9!×2 for two different arrangement for sam and jane. Subtracting it from 10! We get 8×9! C

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Sam and Jane go to a birthday party of a friend. They have an argument 15 Jun 2017, 03:22
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danlew
Sam and Jane meet each other at a party. They have an argument in the party and do not talk to each other for the rest of the party. A photographer wants to take a picture of all the guests at the party. How many options does the photographer have to arrange the 10 guests at the party, if Jane does not want to stand next to Sam when the picture is taken?

A. 9!
B. 8!x9
C. 8x9!
D. \(\frac{10!}{2!}\)
E. 10!

Glue those two together. We'd have 9 units: 1, 2, 3, 4, 5, 6, 7, 8, and {Sam, Jane}. These 9 units can be arranged in 9! ways. Sam and Jane within their unit can be arranged in 2! ways. Thus the total number of ways to arrange 10 people so that Sam and Jane to stand together is 9!*2!.

10 people without any restriction can be arranged in 10! ways, thus in 10! - 9!*2! arrangements Sam and Jane will not be standing together.

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

Answer: C.
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Sam and Jane go to a birthday party of a friend. They have an argument 17 Jun 2017, 05:25
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danlew
Sam and Jane meet each other at a party. They have an argument in the party and do not talk to each other for the rest of the party. A photographer wants to take a picture of all the guests at the party. How many options does the photographer have to arrange the 10 guests at the party, if Jane does not want to stand next to Sam when the picture is taken?

A. 9!
B. 8!x9
C. 8x9!
D. \(\frac{10!}{2!}\)
E. 10!
Show more

We can use the following equation:

(Total number of ways to arrange the guests) = (number of ways with Jane next to Sam) + (number of ways with Jane not next to Sam)

Thus:

(Number of ways with Jane not next to Sam) = (total number of ways to arrange the guests) - (number of ways with Jane next to Sam)

The total number of ways to arrange the 10 guests in a row is 10P10 = 10!.

To determine the number of ways to arrange the guests when Sam and Jane are next to each other, consider Sam and Jane as a single ‘person.’ Thus, it is like placing only 9 individuals in line, and this is done in 9! ways. However, we also have to consider that Sam and Jane, when next to each other, can be ordered either as [S-J] or as [J-S], and so we must multiply the previous value by 2. Thus, the total number of ways to arrange the 10 guests when Sam is next to Jane is 9! x 2.

Thus, the number of ways with Jane not next to Sam = 10! - 9! x 2 = 10 x 9! - 9! x 2 = (10 - 2) x 9! = 8 x 9!.

If the arithmetic in the previous paragraph is confusing, let’s begin with 10! - 9! x 2.

Note that 10! can be re-expressed as 10 x 9!.

10 x 9! - 9! x 2

Rearrange to get 9! x 10 - 9! x 2.

Now pull out the common factor 9! to get 9!(10 - 2) = 9! x 8, or 8 x 9!

Answer: C
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Sam and Jane go to a birthday party of a friend. They have an argument 10 Nov 2025, 05:24
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