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Question

In how many ways, can we get a sum greater than 17 by throwing six distinct dice

A
66( 17C66 11C5)
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B
66( 17C6+6 11C5)
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C
56( 17C66 11C5)
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D
56( 17C6+6 11C5)
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Solution

The correct option is A 66( 17C66 11C5)
Let x1,x2x6 be the number that appears on the six dice.
Let us find the number of ways to get the sum less than or equal to 17.
This will be same as finding the number of solutions to the inequality x1+x2+x3+...x617
Introducing a dummy variable x7 (x70), the inequality becomes an equation :
1+x2+x3+...x6+x7=17
Here ,1xi6 where i=1,2,...6 and x70
Therefore number of solutions
=coeff. of x17 in (x+x2++x6)6×(1+x+x2+)
=coeff. of x11 in (1x6)6(1x)7
=coeff. of x11 in (16x6)(1x)7
= 17C66× 11C5
Total number of cases =66
Hence, the number of ways to get a sum greater than 17=66( 17C66×11C5)

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