There are only two possibilities... 10 groups of 2 or 5 groups of 4. Unless - you can have varying sized groups - which you didn't specify.
No problem has been defined.
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
if order does not matter then, (23x22x21x20x19)/(5x4x3x2x1) = 33,649
There are four different groups of three, each of which can be arranged in six different ways so your answer is 24.
20 x 19 x 18/3 x 2 = 1,140 groups
No problem has been defined.
Combinations of 2: 20*19/(1*2) = 190 Combinations of 3: 20*19*18/(1*2*3) = 1140 Combinations of 4: 20*19*18*17/(1*2*3*4) = 4845 So 6175 in all.
30C8 = 5,852,925
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
if order does not matter then, (23x22x21x20x19)/(5x4x3x2x1) = 33,649
There are four different groups of three, each of which can be arranged in six different ways so your answer is 24.
20 x 19 x 18/3 x 2 = 1,140 groups
To find the number of different groups (a) within a larger group (b), use the formula b!/a!.(b-a)! where the ! sign indicates "factorial". In your problem b = 21 and a = 5 so we have 21!/(5!.16!) this simplifies to 21.20.19.18.17/5.4.3.2 cancelling leaves 21.19.3.17 ie 20349
ya board exam is necessary...Because it ll mainly affect the diploma students....it ll create a problem in choosing different types of groups...
combinations
diagonal
5