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How many ordered pairs (A,B), where A,B are subsets of {1,2,…5} are there, such that |A∩B|=1 ?
i have 40 equal squares in 4 columns and 10 rows. If I chose one square from each column, how many groups of 4 is possible ? I began counting but it got overwhelming so I need to know if there's a formula. The order of the group doesn't need to be consecutive in the sense that if the 4 squares on the first top row is 1.2.3.4 and the 2nd top row squares are 5,6,7,8. Then 1,2,3,8 or 2,8,1,3 is the same group..doesn't matter the order. As long as One is selected from each row, I imagine the amount of groups that is possible is over 10,000
Calculate sum over permutations \tau(i,j): let $x\in R_{2n}$
$\sum_{\tau(i,j)}(x_i-x_j)^2(a(i,j))^2.$

If a(i,j)=1 if j<=n<i; equal to -1 if i<=n<j, equal 0 otherwise.
in a class of 38 children 19 play tennis, 21play squash and ten students do neither how many children play
a)both sports? b)sqaush? c)tennis?
There is a bag of balls 10 in all 5 white and 5 black you take randomly all balls and place them in a row (randomly) watt is the probability to get the 5 white balls on the left and the 5 black wans on the right ?
how many 4 digits numbers can we make using the digits 3,6,7 and 8 without repetitions?
Hi, i have 8 byte series for 1,2,3,4,5,6,...sequentially. and i want to know the function(formula) for solving these series that a particular number i can find from a particular series.
my series is:-

8 byte series finding value
20 95 4a a5 52 a9 18 c0 = 1,
a0 54 2a 95 4a a5 0a c0 = 2,
00 f8 fb fd 7e bf df bf = 3,
c0 56 ab d5 6a b5 02 40 = 4,
00 16 8b c5 62 b1 14 40 = 5

so, what should be the function(formula) for these series . that i can apply all the series and i can find the number according to that series.
I know the correct workings. But I need to know why the ones I did won't give me the correct answer. Like, what really is wrong with them. I am following logic. Where do I fail??

In how many ways can a committee of 4 men and 4 women be seated in a row if no one is seated next to a person of the same gender?

I do it like this:

there are 4 men and 5 possible positions for the women to be seated therefore. the first woman can choose between any of the 5, the other three cannot occupy the positions at the ends of the row since that would cause two of the men to be sitting together. Hence there are only 3 possible seats for each of the three women, then 2 for the remaining two women and then 1 for the last. I multiply 4! by 5*3*2*1 therefore. The answer is not this however.

What did I do incorrectly?

It's a method in a Cambridge endorsed book that I use. But obviously I am doing something wrong, according to the author and yahooanswers.com. What do I do??????????
Two floors of a store are connected by two up and down escalators
Floor 1 has 7 people& floor 2 has 12 people
There are always two people occupying each escalator
Question 1 :
If there are always 2 people on each escalator at any one moment, will the amount of people on each floor ever change?
Question 2 :
If there are always 2 people on each escalator at any one moment, will the specific people on each floor ever change ?
what is the remainder when 7^7^7^upto infinite times(7 to the power 7 to the power 7 to the power ....infinite times) is divided by 13
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