Maximim number of triggers that can be applied to a single table is 12

Silverlight 1 is purely AJAX and JavaScript based. All the code has to be written in JavaScript and XAML. 

 Silverlight 2 supports managed code. When the Silverlight 2 runtime is installed, it installs a limited version of the .NET runtime on the client machine

Possibilities     Bowlers      Batsmen         Number of ways

                         6               9

         1              4                7              (6C4*9C7)

         2              5                6              6C5*9C6

         3              6                5              6C6*9C5

6C4*9C7 = 15 x 36 = 540

6C5*9C6 = 6 x 84 = 504

 6C6*9C5= 1 x 126 = 126

Total = 1170

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = 7C3 = 35

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

Selecting zero'A's= 1

Selecting one 'A's = 1

Selecting two 'A's = 1

Selecting three 'A's = 1

Selecting four 'A's = 1

Selecting five 'A's = 1

=> Required number ofways =6

Total number of alphabets = 10

so ways to arrange them = 10! 

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps 

Hence number of arrangements = 10!/4! x 3! x 2! = 12600

No.Minimal spanning tree assures that the total weight of the tree is kept at its minimum. But it doesn’t mean that the distance between any two nodes involved in the minimum-spanning tree is minimum.