I was asked to solve the following problem. A triangle has sides of length 12,8 and 6. There are line segments of equal length that are parallel to each side of the triangle and intersect at one point(concurrent) inside the triangle, what is the length of line segment? I solved the problem by...
If the prime factorization of C+1=(P1)^(k1)*(P2)^(k2)*...*(Pn)^(kn) then the number of divisors of C+1 equals (k1+1)*(k2+1)*...*(kn+1). The number of paired solutions becomes the smallest integer >=[(k1+)*(k2+1)*...*(kn+1)]/2 . Take for example x+y+xy=(3^4)*(5^2)*(7^3)-1 then C+1=...
Using your example I think this will help:(x+y)+xy+1=6496+1 then (1+x)(1+y)=6497. The number of solutions can then be determined from the prime factorization of 6497.
x=72 and y=88
You can prove Sum(nCk,k=0,1..n)=2^n as a straight summation problem by induction on n. It works for n=0 then assume it is true for n-1. Since 2^(n-1)+2^(n-1) = 2^n expand out the summations and rearrange terms and show (n-1)C(k-1)+(n-1)Ck=nCk Where nCk=n!/k!/(n-k)! and note that...
Yes, except P(Y1>=0.5| Y2<=0.25)= D/(D+B)=3/7. The upper limit of integration in the denominator of your 2nd problem should be 3/4. xs=3/4 and ys=1/4 and
P(Y1>=0.5|Y2=0.25)=(2*ys)/(2*xs)=1/3. Nice diagram.
If you draw the triangle then the limits are simple. Your first problem becomes a ratio of areas and the second problem becomes a ratio of line segments. You will need to change the upper limit of the denominator of the second problem.
Hi, maybe this will help. Using n!=n*(n-1)! consider this sequence
3!=3*2!
2!=2*1!
1!=1*0!
Using the last sentence, the left hand side has 1! which equals 1 and the right hand side has the product of 1 * something we call 0!. So 0!=1.
Probability of at least one ace is ((4C1 * 46C2)+(4C2 * 46C1)+(4C3 * 46C0))/(50C3)=0.225510204082.
Probability of no aces is (46C3)/(50C3)=0.77448979597837. Where nCr=n!/r!/(n-r)!
Start with an easy sequence {2,4,6,...} and have the children give you the Nth term. Afterward, you might try the secret number approach. Write X on the board and ask if anyone can guess what is your secret number. Ask if they need more of a hint then depending on the skills of the children...
The probability that N people have different birthdays is (365/365)*(364/365)*(363/365)*...*((366-N)/365) assuming that the events are independent of each other.
A leap year calc is (366/366)*(365/366)*(364/366)*...*((367-N)/366) for N people.
Wouldn't you count the number of ways n people could have different birthdays as
365!/(365-n)!=365*364*363*...*(365-n+1). First person can be born on any date (365 ways) in the year, second person must be born on a different date ( 364 ways ) , third person must be born on a date different that...