There is an octagonal die with 8 faces having values 1 to 8 on each face. It is thrown 4 times in a sequence. In how many ways can you get a sum which is less than or equal to 20?

1245

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1395

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1575

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2865

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Solution

The correct option is D
2865

This is a question based on both upper limit and lower limit

Given that A+B+C+D $\leq$ 20

This can be written as A+B+C+D+E=20

Where 1 $\leq$ A,B,C,D $\leq$ 8

Applying a lower limit of 1 to A,B,C,D

A+B+C+D+E=16

$N u m b e r o f w a y s =^{20} C_{4} = 4845$

Giving an upper limit of 8 to A (this makes A=9)

A+B+C+D+E=8

Number of solutions = $^{12}$ C $_{4}$

Similarly, violating the conditions for A,B,C & D= 4* $^{12}$ C $_{4}$

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