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A staircase is to be constructed between M and N: figure.

The distance from M to L is 5 meters and the distance from L to N is 2 meters. If the height of a step is 25 centimeters and its width can be any integer multiple of 50 centimeters, how many different ways can we arrange the number and width of the steps?

Since the height is locked in at .25 m and the height of the staircase is 2 m, there has to be 8 steps. Since the width of each step is an integer multiple of .5 m and the total run of the staircase is 5 m, then there are 5 m/.5 m = 10 possible arrangements for the width of the steps. So,n = 10 and r = 8 since there are 10 possible arrangements for the width but you have to choose 8 for the 8 steps. So, I did a combination with n = 10 and r = 8 since the order of the steps is irrelevant and I got a total of 45 arrangements for the width of the steps.But the answer at the back of the book says 36. Can someone point out the mistake? Thanks!

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I think the difference between your answer and the book is that you are allowing the staircase to start later than $M$ - effectively to put the extra distance, the two "spare" $0.5$m increments, before the first riser. That gives the $45$ options answer. If you constrain the staircase to have the first riser at $M$, you get the result of $36$ options.

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As according to height we need 8 stairs. And according to width we have 10. But if we multiply 10 with height .25 we get 2.5 m, which is not true as we have height 2 m. So it means two stairs with width 1 m each or one with width 1.5 m and another .5 m.

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    In fact, you also need to include the option that one stair tread is $1.5$m wide. – 2017-01-29
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    Yes thank you. I skip it. – 2017-01-29