Question

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?

Answer 1

• 46

Step-by-step explanation:

1. Number of sequences with exactly seven heads:

• First head can go in 9 different positions
• Second head can go in 8 different positions
• Third head can go in 7 different positions

• Fourth head can go in 6 different positions
• Fifth head can go in 5 different positions
• Sixth head can go in 4 different positions
• Seventh head can go in 3 different positions

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.

• That is: 9 × 8 / 2 = 36 different ways of tossing seven heads

2. Number of sequences with exactly eight heads:

Following the same reasoning:

• 9×8×7×6×5×4×3×2 divided by 8×7×6×5×4×3×2×1 = 9

• That is 9 different ways of tossing eight heads.

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:

• 36 + 9 + 1 = 46 ← answer