you have 15 places, for one sister can be. for the other sister, there are 14 more places, and given that sister A isnt on the edge, 2 places out of the 14 she could be in, if A is on the edge, there is only 1 out of 14 places, so the answer is
P(0<A<15) = 13/15, P(B next to A | 0<A<15) = 2/14
P(A=0 or A=15) = 2/15, P(B next to A | A=0 or A=15) = 1/14
P(B next to A) = 13/15*2/14 + 2/15*1/14 = 28/210 = 2/15 = 0.1333...
I come up with a different solution.
A: Consider all of the different pairings the 2 could be in. There are (15 choose 2)possibilities. That's 105 different pairings. Now count how many pairings are next to each other: (1,2), (2,1), (2,3), (3,2), ... (14,15) and (15, 14). That's 28 pairings out of the possible 105. That's .2666, which is double Luca's answer. And that explains my mistake. There are 15 * 14 possible pairings, not (15 choose 2). So that makes it 28 possibilities out of 210 which is .133333.
B: You still have the 210 different pairings but the number of pairings that meet the requirements is smaller: (1,7), (7,1), (2, 8), (8,2),..., (9,15) & (15,9). There are 18 good pairings out of the 210, or .0857
2. I have no idea what you're asking for here.
Thank you,and the second problem is a combinations with repetition problem.
1.At tht Mary sorority the 15 sisters who are seniors line up in a random manner for a graduation picture.
Two of these sisters are Columba and Piret.What is the probability that this graduation picture will find
(a)Piret and Columba standing next to each other?(b)exact five sisters standing between Columba and
2. Consider the 2^19 compositions of 20.(a)How many have each summand
even?(b)How many have each summand a multiple of 4?
Thank you very much~~:cool:
Something is wrong in b.
If she has 2 dozen of each (24), and she is only picking 20 beads total, then it doesn't matter how many of each she has. It could be infinite.
So she has n choices for the first bead, n choices for the second bead, n choices for the.... which is n*n*n....*n = n^20. So n^20 = 230230 which comes out to 1.85399, which must be wrong.
Unless I'm missing something.
the ans is n=7.
a) This is a combination problem. Basically, you are choosing three vertices from a set of n, where order doesn't matter. So, the formula is:
This will give you the number of triangles you can make from the vertices of any regular polygon!
b) In order for this to work, no two vertices may be adjacent to another. Unfortunately, I don't have time right now to figure out that formula.