We have 20 people of team A ,20 people of team B and 20 double rooms .How we can put the peaople so that in the rooms we only have people from the same team.
How many ways can 10 rooms be chosen from 20 rooms
20C10 = 184756
Team A can have these or they can have the other set. So that is 2 ways.
Now how many ways can I pair off 20 people?
Person 1 can go with anyone, 19 to chose from
Now there are 18 people left
Person 3 can go with anyone 17 to chose from
so there will be 16 left
and so on.
So that seem to be 19*17*15*13*11*9*7*5*3*1 = = 654729075
I really do not know if that is correct.
So I have
184756*2*654729075*654729075 158398768771746207705000 = 1.5839876877174621 * 10^23
Mmm that is a lot!
That is my best guess. (For the moment)
The digits 1 up to 4 are written on a spinning top. Ella spins the top five times. This way she'll get a number of five digits, like 42313 or 13421
a. How many different numbers can Ella get? 45 = 1024 numbers
4^5=1024 different possibilities Correct
b. How many different numbers can she get if the number starts with an even digit and then alternates with an odd digit and then an even digit. 25 = 32
even, odd, even, odd, even, 2^5=32 correct
c. How many different numbers can she get if she spins two times a 2, two times a 4 and then one other (1 or 3) digit?
2,2,4,4, (1 other 2 to chose from)
22441, or 22443 in any order
60 possibilities (I'm pretty sure)
The range is the y values, The highest y value (not actually included) is 4. The curled brackets indicates that neither infinity or 4 are actually included.
hene the range is (-infinity, 4)
The arrows are suppose to indicate the direction that the graph will take from that point on.
Hence on the left it goes down for ever and on the right it goes up forever.
Hence the range is (-infinity, +infinity.)
At a birthday party, 5 boys and 3 girls are seated around a table.
How many different arrangements are possible if no two girls are seated next to each other?
Normally for these questions rotations are considered to be the same.
Reflections are considered to be different
Are you interested in gender arangements or individual people arrangements?
If you are only interested in gender positions then.
Seat a girl first. She is like the begining of the row.
Now there is 5 boys
We need to slot in 2 more girls
The stars indicate where the other 2 girls can go.
4C2 = 6 gender arrangements are possible.
If you are interested in the position of individuals then
Stil sit any girl first. (could have chosen a boy if you wanted to)
Now there is 5 boys and 2 girls left to seat.
Sit the 5 boys first. This can be done in 5! = 120 ways
So we have GBBBBB and we have to sit 2 more girls.
4C2*2 = 12 ways.
So that is 120*12 = 1440 individual child arrangements.
Rocco is a member of the Aviation Club, which has 12 members.
The members need to choose a 3-person Helicopter Committee and a 4-person Glider Committee.
A member can serve on one of the committees or both. Rocco refuses to serve on both (but he can serve on one of the committees).
In how many ways can they form the committees?
If rocco is on the helicopter committee then
there are 11C2*11C4 = 55*330 = 18150 ways
If rocco is on the Glider committee committee then
there are 11C3*11C3 = 165*165 = 27225 ways
If rocco is not on either committee then
there are 11C3*11C3 = 165*330 = 54450 ways
18150+27225+54450 = 99825 ways