A certain company assigns employees to offices in such a way that some of the offices
can be empty and more than one employee can be assigned to an office. In how many
ways can the company assign 3 employees to 2 different offices?
A. 5
B. 6
C. 7
D. 8
E. 9
On a map Town G is 10 centimeters due east of Town H and 8 centimeters due south of
Town J. Which of the following is closest to the straightline distance, in centimeters,
between Town H and Town J on the map?
6
13
18
20
24
The second problem looks simple but i could not 13 as an answer.
please help
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Lets say the two offices are named A & B. Possible no. of ways of assigning 2 offices to three people are:deepakb wrote:A certain company assigns employees to offices in such a way that some of the offices
can be empty and more than one employee can be assigned to an office. In how many
ways can the company assign 3 employees to 2 different offices?
A. 5
B. 6
C. 7
D. 8
E. 9
 0 in A and 3 in B = 1 way
 3 in A and 0 in B = 1 way
 2 in A and 1 in B = 3C2 = 3 ways
 1 in A and 2 in B = 3C2 = 3 ways
Total no. of ways = 1+1+3+3 = 8 ways  Answer (D)
Does this look right?
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Q1:
Assume the offices are 1 and 2, and employees are A, B, C. define your different scenarios, find how many combinations are there for each scenario, then add. The scenarios are:
all three employees in 1, (and none in 2).
two employees in 1 (and the last one in 2)
one employee in 1 (and the last two in 2)
no employees in 1 (all three in 2).
Here are the formulas for the number of combinations for each scenario:
scenario (3  0) only one option, or 3C3=1
scenario (2  1) 3C2 = 3!/2!1! = 3
scenario (1  2) 3C1, which is the same as 3C2 = 3!/1!2!=3
scenario (0 3) only one option, or 3C0 = 3!/3!0! = 1.
total of [spoiler]1+3+3+1=8[/spoiler]
There's an easier way to look at the problem, but it's more difficult to conceptualize: there are 3 employees, and each employee has 2 choices for a room.
Therefore, there are
2 choices for employee A and
2 choices for employee B and
2 choices for employee C
each "and" denotes "multiply", so total of [spoiler]2*2*2=8[/spoiler] ways of assigning employees. The benefit of this approach is that it is scalable: 3 employees with 3 rooms would have [spoiler]3*3*3=27 ways[/spoiler].
Q2: Use the Pythagorean theorem to find the hypotenuse of the resulting triangle:
8^2+10^2 = x^2
64+100 = 164 = x^2
The closest perfect square is 13^2=169, so 13 is the closest answer.
Assume the offices are 1 and 2, and employees are A, B, C. define your different scenarios, find how many combinations are there for each scenario, then add. The scenarios are:
all three employees in 1, (and none in 2).
two employees in 1 (and the last one in 2)
one employee in 1 (and the last two in 2)
no employees in 1 (all three in 2).
Here are the formulas for the number of combinations for each scenario:
scenario (3  0) only one option, or 3C3=1
scenario (2  1) 3C2 = 3!/2!1! = 3
scenario (1  2) 3C1, which is the same as 3C2 = 3!/1!2!=3
scenario (0 3) only one option, or 3C0 = 3!/3!0! = 1.
total of [spoiler]1+3+3+1=8[/spoiler]
There's an easier way to look at the problem, but it's more difficult to conceptualize: there are 3 employees, and each employee has 2 choices for a room.
Therefore, there are
2 choices for employee A and
2 choices for employee B and
2 choices for employee C
each "and" denotes "multiply", so total of [spoiler]2*2*2=8[/spoiler] ways of assigning employees. The benefit of this approach is that it is scalable: 3 employees with 3 rooms would have [spoiler]3*3*3=27 ways[/spoiler].
Q2: Use the Pythagorean theorem to find the hypotenuse of the resulting triangle:
8^2+10^2 = x^2
64+100 = 164 = x^2
The closest perfect square is 13^2=169, so 13 is the closest answer.
thank u.........Geva Stern wrote:Q1:
Assume the offices are 1 and 2, and employees are A, B, C. define your different scenarios, find how many combinations are there for each scenario, then add. The scenarios are:
all three employees in 1, (and none in 2).
two employees in 1 (and the last one in 2)
one employee in 1 (and the last two in 2)
no employees in 1 (all three in 2).
Here are the formulas for the number of combinations for each scenario:
scenario (3  0) only one option, or 3C3=1
scenario (2  1) 3C2 = 3!/2!1! = 3
scenario (1  2) 3C1, which is the same as 3C2 = 3!/1!2!=3
scenario (0 3) only one option, or 3C0 = 3!/3!0! = 1.
total of [spoiler]1+3+3+1=8[/spoiler]
There's an easier way to look at the problem, but it's more difficult to conceptualize: there are 3 employees, and each employee has 2 choices for a room.
Therefore, there are
2 choices for employee A and
2 choices for employee B and
2 choices for employee C
each "and" denotes "multiply", so total of [spoiler]2*2*2=8[/spoiler] ways of assigning employees. The benefit of this approach is that it is scalable: 3 employees with 3 rooms would have [spoiler]3*3*3=27 ways[/spoiler].
Q2: Use the Pythagorean theorem to find the hypotenuse of the resulting triangle:
8^2+10^2 = x^2
64+100 = 164 = x^2
The closest perfect square is 13^2=169, so 13 is the closest answer.