Question

number of ways in which score of 11 can be made from a throw of 3 persons each throwing a single die once?

Answer 1

ok so what i did is x1 + x2 +x3 = 11, number of solutions=13c2
then we have to subtract, 3*(4c2) --> refer http://openstudy.com/users/yrelhan4#/updates/51320a88e4b0f0611bc045e6
but i am not getting the answer.
where am i going wrong?

Answer 2

correction: 3*(5c2)

Answer 3

@shubhamsrg

Answer 4

641, 632,
551, 542, 533,
443,

Answer 5

Kinda tedious but you could do it by hand... to get 11 (hopefully i didn't miss any):

6, 4, 1
6, 3, 2
5, 5, 1
5, 4, 2
5, 3, 3
4, 4, 3

Answer 6

are you sure you have to do it by hand? answer is 27. :O

Answer 7

Then you might have to account for the fact that each of those can happen in 3P3 ways, if it matters who throws what number.

Answer 8

Then you'll have to subtract a couple, since 5,3,3 is the same as 5,3,3

Answer 9

the sets with three different numbers has 6 possible orders,
the sets with two different numbers have 3 possible orders,
the sets with one repeated number have only one order .

6+6+3+6+3+3=

Answer 10

hmmm. thank you. :)

Answer 11

It seems difficult to do with just straight formulas, given that your x1 x2 x3 values are limited between 1 and 6.

Answer 12

i know. but we did a problem yesterday with those formulas. i think i am doing something wrong. i'll just give it a try and post the work if i get it.

Answer 13

6, 4, 1 .... 3P3
6, 3, 2 .... 3P3
5, 5, 1 .... (3P3)/2
5, 4, 2 .... 3P3
5, 3, 3 .... (3P3)/2
4, 4, 3 .... (3P3)/2

so 3*(3P3) + 3*(3P3)/2

Answer 14

^ hmm. i'll note down that method too. :)

Answer 15

ok i got what i was looking for.
first since x,y,z>=1, the equation will be x+y+z=7..
number of solutions = n-1Cr-1=9C2
now referring to the method in the link i posted before (i dont know how to explain it), we have to subtract 3*(4C2). hence the answer.

Answer 16

thank you everybody, i appreciate. :)

Answer 17

for that method, yep, we'll first account for 6>=x>=1 , 6>=y>=1 and 6>=z>=1,

or
5>=x>=0 , 5>=y>=0 ,5>=z>=0

=> (5-x) + (5-y) + (5-z) = 7

=> C(9,2) = 36

from here, we will subtract cases when our X>=6, Y>=6 and Z>=6

=> 3* C(3,2) = 9
-> ans= 36-9 = 27

Answer 18

make that 5>=X>=0 , 5>=Y>=0 ,5>=Z>=0

where X=x-1
Y=y-1
and Z=z-1

Answer 19

yup.. i was a little wrong there.