Number of ways to choose an order pair (a,b) of numbers from the set {1,2,3..........10} such that |a-b|≤5 is

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100

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Solution

The correct option is C
80

$| a - b | \leq 5$ means| a-b|=0 or $1 \leq | a - b | \leq 5 | a - b | = 0 \Rightarrow a = b$ which is possible in 10 ways. Let x, y, z represent, numbers before, in between and after a and b then $x + y + z = 8, 0 \leq y \leq 4$ and $x, z \geq 0$

= coefficient of $x^{8} i n (1 + x + . . . . x^{8})^{2} (1 + x + . . . . . x^{4})$

= 35

a, b, can be in any order

Total ways = $2 \times 35 + 10 = 80.$

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