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if 4 dice are rolled, then no of ways getting sum of 10

a)56
b)64

c)72

d)80
| 32 views

+1 vote

We can solve it by star and bar method.

Suppose $1$st dice score is $x_1.$

Suppose $2$nd dice score is $x_2.$

Suppose $3$rd dice score is $x_3.$

Suppose $4$th dice score is $x_4.$

We want $x_1+x_2+x_3+x_4 = 10$ with constraints $1\leq x_1\leq 6$ , $1\leq x_2\leq 6$ , $1\leq x_3\leq 6$,$1\leq x_4\leq 6$

So $x_1,x_2,x_3,x_4$ have minimum values $1$ and remaining $6$ can be scored as $x_1+x_2+x_3+x_4 = 6$

Total no. of ways to score $6=\binom{6+4-1}{6} = \binom{9}{6}= \binom{9}{3} = \frac{9*8*7}{3*2*1}=12*7=84$

Now when we are distributing $6$ among $x_1,x_2,x_3,x_4$ we can’t have cases $(6,0,0,0),(0,6,0,0),(0,0,6,0),(0,0,0,6)$ since otherwise constraints $1\leq x_1\leq 6$ , $1\leq x_2\leq 6$ , $1\leq x_3\leq 6$,$1\leq x_4\leq 6$ will not be satisfied.

$\therefore$ answer = $84-4=80$

Option $D$ is correct.

see this it is useful to read https://brilliant.org/wiki/integer-equations-star-and-bars/

by (4.1k points)
0
How you take care of upper bound in this example ?
0
here upper bound cases would not fit into solution, so they would be handled automatically.

Like x1 =7 then other x can't be 0

and the ones that were fitting but now following lower bound constraints, i removed them.
0
Ok... Fortunately its sum need to be 6, so it's worked !
0
yes correct.