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5. How many different ways are there to choose 6 out of 10 books if the order does not matter?​

Sagot :

Kristofer55

Since the order of the books matter, the first book can be one of 10 choices. The second can be one of 9 choices, etc.

In the end, if you have n books and k can fit on a shelf then there are n!(n−k)! ways to arrange them.

With n=10 and k=6 we get 10×9×8×7×6×5=15120

angelcidamon

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This is the problem of permutation and combination

So, there are 6 places for 10 books on the shelf

the first place can be filled by any one of 10 books. now out of 10 books 1 is already filled, so we have 9 books and 5 places.

the second place can be filled by anyone of 9 books. Now we have 8 books and 4 places

similarly, for 3rd place, we have 8 choices of books, and 7 books are remaining with 3 places, and

for 4th place→ 7 choices

for 5th place → 6 choices

for 6th place → 5 choices

ANS: Therefore the number of ways 10 books can be fit in 6 places is 10*9*8*7*6*5 = 1,51,200.

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