<span>Assuming that "pair up students" means "divide up all 20 of the students into groups of two," and we regard two pairings as the same if and only if, in each pairing, each student has the same buddy, then I believe that your answer of 20! / [(2!)^10 * (10!)] is correct. (And I also believe that this is the best interpretation of the problem as you've stated it.)
There are at least two ways to see this (possibly more).
One way is to note that, first, we have to select 2 students for the first pair; that's C(20, 2) (where by C(20, 2) I mean "20 choose 2"; that is, 20! / (18! * 2!). )
Then, for each way of selecting 2 students for the first pair, I have to select 2 of the remaining 18 students for the second pair, so I multiply by C(18, 2).
Continuing in this manner, I get C(20, 2) * C(18, 2) * ... * C(2, 2).
But it doesn't matter in this situation the order in which I pick the pairs of students. Since there are 10! different orders in which I could pick the individual pairs, then I want to divide the above by 10!, giving me the answer
[C(20, 2) * C(18, 2) * ... * C(2, 2)] / 10!.
This is the same as your answer, because C(n , 2) = n(n - 1) / 2, so we can simplify the above as
[(20 * 19) / 2 * (18 * 17) / 2 * ... * (2 * 1) / 2] / 10!
= 20! / [2^10 * 10!]
= 20! / [(2!)^10 * (10!)].
Another way is to reason as follows:
1. First, arrange the 20 students in a line; there are 20! ways to do this
2. We can get a pairing by pairing the 1st and 2nd students in line together, the 3rd and 4th students together, etc.
3. But if I switch the order of the 1st and 2nd student, then this doesn't give a different pairing. I don't want to count both orderings separately, so I divide by 2!
4. The same argument from step 3 holds for the 3rd and 4th student, the 5th and 6th student, etc., so I need to divide by 2! nine more times
5. Finally, the particular order in which I selected the ten pairings are unimportant--for example, the following orderings don't produce different pairings:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
3, 4, 1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
So I need to further divide by the number of ways I can arrange the ten pairs, which is 10!.
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Answer:
the ans is :
Step-by-step explanation:
First, it would be helpful to draw a quick sketch of the lines. It helps to visualize the problem.
To find the intersection point, we need to find the point where x and y are the same value in both equations.
The line equations:
6x+2y=26 ................... 1
2x+3y=18 ................... 2
Can be rearranged to the common line equation form: y = mx + c
y = 13 - 3x ................... 3
y = 6 - 2/3 x ................. 4
At the intersection point, y will be equal for both equations. So, we can set 3 equal to 4 and solve for x.
13- 3x = 6 - 2/3 x
13 = 6 + 3x - 2/3x ....... add 3x to both sides
13 = 6 + 2 1/3x ........ simplify
7 = 2 1/3 x ........ subtract 13 from both sides
7 = 7/3 x ......... multiply both sides by 3/7
3 = x
To calculate the y-coordinate substitute x = 3 into 3.
y = 13 - 3x
y = 13 - 3(3)
y = 4
To check your answer, substitute the values for x and y into the other equation, 4.
The point of intersection is (3,4).
If you drew a sketch of the problem, you should be able to see that this point of intersection makes sense.
Answer:
Cost of shipping and handling = $23.453
Step-by-step explanation:
Given
Price of Chair=$220.59
Tax=7.9%
Invoice Price=$261.47
In order to find the cost of shipping and handling, we have to subtract the cost of chair and the tax from the invoice price of chair.
To find the tax,
Tax amount=220.59*0.079
=$17.42661
Now,
Cost of shipping and handling=$261.47-$220.59-$17.42661
=$23.453 ..
