Solution to the Coin Problem
In the description of the problem , we saw that when the coin is flipped once, there are two possible outcomes:
1 H or 1 T
When there are two flips of the coin, there are four possible outcomes:
1 HH
2 with 1 H and 1 T (1 HT and 1 TH)
1 TT.
Consider the possible outcomes when the coin is flipped 3 times:
| Number of Heads and Tails |
Description | Possible Outcomes |
Number |
|---|---|---|---|
| 3 Heads | HHH |  |
1 |
| 2 Heads 1 Tail |
HHT HTH THH |
   |
3 |
| 1 Head 2 Tails |
TTH THT HTT |
   |
3 |
| 3 Tails | TTT |  |
1 |
Notice that we get:
1 with 3 H (HHH)
3 with 2 H and 1 T (HHT, HTH, and THH)
3 with 1 H and 2 T (TTH, THT, and HTT)
1 with 3 T (TTT)
Compare these outcomes with the numbers in Pascal's triangle:
| 1 | ||||||||||||||
| 1 | 1 | |||||||||||||
| 1 | 2 | 1 | ||||||||||||
| 1 | 3 | 3 | 1 | |||||||||||
| 1 | 4 | 6 | 4 | 1 | ||||||||||
| 1 | 5 | 10 | 10 | 5 | 1 | |||||||||
| 1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||||||
| 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 |
The outcomes correspond to the first few rows of Pascal's Triangle. The entries in each row of Pascal's Triangle tell us the number of different ways a particular combination of Heads and Tails occurs. For example the numbers in the 4th row tell us we would have:
1 with 4 H
4 with 3 H and 1 T
6 with 2 H and 2 T
4 with 1 H and 3 T
1 with 4 T
The problem asks us how many different ways we can get 4 Tails and 3 Heads with 7 flips of the coin. To find the answer look at Row 7 of Pascal's Triangle:
| 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 |
There will be:
1 with 7 H
7 with 6 H and 1 T
21 with 5 H and 2 T
35 with 4 H and 3 T
35 with 3 H and 4 T
21 with 2 H and 5 T
7 with 1 H and 6 T
1 with 7 T
Therefore there will be 35 different results with 4 Tails and 3 Heads that are possible when a coin is flipped 7 times.
Try another problem that uses Pascal's Triangle in the solution.
email the author: Bruce Jacobs
Last modified: Thursday, February 6, 2003