# Search results

1. ### Probability solve the expression P(2n+4,3) = 2/3P(n+4,4) for n

this wat i did... (2n+4)!/((2n+4)-3)! = (2/3(n+4)!)/((n+4)-4)! (2n+4)!/(2n+1)! = (2/3(n+4)!)/(n)! i dont know wat do do next, im soo stuck, since you cannot cancel any of them, so should i cross multiply it?
2. ### Probability solve the expression P(2n+4,3) = 2/3P(n+4,4) for n

permutations ok, i get what you mean, but how can i manipulate this, is seems i cannot cancel it... help again
3. ### Baseball permutations help

so the answer would be 9(players) * 8(positions) * 1(nineth - pitcher will be the last) = 72
4. ### Baseball permutations help

i cannot get the question since i dont know how to play baseball... this is the question... The manager of a baseball team has picked the nine players for the starting lineup. In how many ways canhe set the batting order so that the pitcher bats last?
5. ### Probability solve the expression P(2n+4,3) = 2/3P(n+4,4) for n

i just want to confirm if my anser is right.... this is the problem: solve the expression for n Є N P(2n + 4, 3) = 2/3P(n+4, 4) this is my work: (2n +4)!/(n-3)! = (2/3(n+4)!)/(n-4)! -2(n-2)!/(n-3)! = -2/3(n-4)!/(n-4)! -2/(n-3)! = -2/3 -2 = -2/3(n-3) -2 = -2n/3 +6/3 -2 = -2n/3...