answer:This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem: How many ways are there to choose k elements from an ordered n element set without choosing two consecutive members? You want to choose k numbers out of n with no consecutive numbers. For each configuration, we can subtract i-1 from the i-th element in your subset. This converts your configuration into a configuration with k elements where the largest possible element is n-k+1, with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection. Without consideration of the second condition, we have: {15 choose 1} + {14 choose 2} + {13 choose 3} + ... + {9 choose 7} + {8 choose 8} Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than k, which translates to subtracting k - 1 from the "top" of each binomial coefficient. Now we have, after we cancel all the terms {n choose k} where n < k, {15 choose 1} + {13 choose 2} + {11 choose 3} + {9 choose 4} + {7 choose 5}= 15 + 78 + 165 + 126 + 21 = 405

answer:There are a total of 5 times 5 = 25 possibilities. Multiplying 1 or 2 by any of the other numbers in the bowl will not result in a number greater than 10, so we know that Josh does not draw 1 or 2. Therefore, Josh must draw a 4 in order for the result to be even. Thus, his possibilities are: (3,4);(4,3);(4,4);(4,5);(5,4), making for 5 possibilities, and a probability of frac{5}{25} = frac{1}{5}.

answer:I need to count the number of ways to choose 6 starters out of 14 players, given that 3 of them are fixed. I can think of this as choosing 3 more starters out of the remaining 11 players, since the triplets are already in. This is a combination problem, because the order of the starters does not matter. I recall that the formula for combinations is nCr = n!/(r!(n-r)!), where n is the total number of elements and r is the number of elements chosen. In this case, n = 11 and r = 3, so I can plug them into the formula. I get 11C3 = 11!/(3!(11-3)!) = (11*10*9*8!)/(3*2*1*8!) = (11*10*9)/(3*2*1) = 165. Therefore, there are 165 ways to choose 6 starters if all three triplets are in the starting lineup.

answer:I notice that this expression has a familiar pattern: it is the expansion of (9 + 1)^3 by the binomial formula. So, I can rewrite it as (9 + 1)^3 and simplify it. This gives me 10^3, which is just 1000.