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**.
 
						

							
Dear student 
Please refer the explanation below 
First selecting five matches , whose predictions can be anything gives a count of 20C5.
Second for these matches, the total options are 3^5.
For the rest , the prediction has to be correct so total options are 1^5=1.
So final answer is (20C5)*(3^5).
P.S These questions are best suited for mathstackexchange, you could get better answers with more explanation
Hope this helps.
Good Luck