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Question

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?

A

1245

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B

1395

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C

1575

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D

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≤20 This can be written as A+B+C+D+E=20 Where 1≤A,B,C,D≤8 Applying a lower limit of 1 to A,B,C,D A+B+C+D+E=16 Number of ways = 20C4 = 4845 Giving an upper limit of 8 to A (this makes A=9) A+B+C+D+E=8 Number of solutions = 12C4 Similarly, violating the conditions for A,B,C & D= 4*12C4

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