Awesome q2a theme

+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 canâ€™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/

- All categories
- General Aptitude 21
- Engineering Mathematics 95
- Digital Logic 59
- Programming & DS 147
- Algorithms 108
- Theory of Computation 144
- Compiler Design 35
- Databases 50
- CO & Architecture 62
- Computer Networks 80
- Non GATE 0
- Others 44
- Admissions 10
- Exam Queries 10
- Tier 1 Placement Questions 2
- Job Queries 1
- Projects 0

932 questions

596 answers

1,885 comments

81,474 users