The results of 20 chess games (win , lose or draw) have to be predicted. How many different forecasts can contain exactly 15 correct results?

For a correct forecast with **exactly** 15 correct results, 5 other must be wrong. For any game to be wrong there exist **2** ways, as in all, there are **3** possible outcome of a game. So we total **(3–1)^5** possible combinations for those *wrongly* predicted games. Hence *for any one* of the possibly “**15-correct”** forecast, we have in all **32** predictions with a similar end-effect.

Now, how many of those **15-correct** forecasts are possible? 20C15 = 20C5 = **15504**, i.e. the number of ways to select those 15 games for which the forecast is correct. Hence in all, we have 15504*32 = **496128**.

 

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Please refer the explanation below