A coin is tossed $9$ times, and at least $7$ of the tosses were heads. How many different sequences of tosses could there have been?

Explanation:

1. Number of sequences with exactly seven heads:

That gives: 9×8×7×6×5×4×3

Since all the heads are equivalent, you have repetitions that are not different. Then, you must discount the combinations that are equivalent.

In how many ways seven heads can be arranged? 7×6×5×4×3×2×1

Then, you must divide 9×8×7×6×5×4×3 by 7×6×5×4×3×2×1.

2. Number of sequences with exactly eight heads:

Following the same reasoning:

3. Number of sequences with exactly nine heads

That is only one way: when all the tosses are heads: 1

4. Number of sequences with at least seven heads

Add the combinations of having exactly seven heads, eight heads and nine heads:

a

step-by-step explanation:

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068