In how many ways can we pick three different numbers out of the group 1, 2, 3, 100 such that the largest number is larger than the product of the two smaller ones? (the order in which we pick the numbers does not matter.)

i will bite - this is not a complete ans but an approximation:

there are 100*99*98 ways to pick 3 different no. out of 100

if 2 of the no. picked are >10, then there is no way the third no. can be larger than the products of the other 2 nos.

there are 90*89*98 ways to pick 2 nos between 10 and 100 and a third no. between 1 and 100.

So an approximate ans = 100*99*98 - 90*89*98 = 185220 but the actual ans will be smaller than this.

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2