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.
20 x 19 x 18/3 x 2 = 1,140 groups
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.
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
20 x 19 x 18/3 x 2 = 1,140 groups
There are four different groups of three, each of which can be arranged in six different ways so your answer is 24.
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
Many different groups will come up with plans to fix the problem
Many different groups will come up with plans to fix the problem