Question

How many ways can a person toss a coin 8 times so that the number of tails is between 5 and 7 inclusive?

Answer 1

We can evaluate case by case from 5 tails to 7 tails. As both cases are included we have:

5 tails out of 8

6 tails out of 8

7 tails out of 8

The number of ways we can have 5 tails trying 8 times is just 8 combined 5. We use combination since the order does not matter here.

Recalling the formula for combination:

`nCk=(n!)/(k!(n-k)!)`

Then:

`8C5=(8!)/(5!(8-5)!)`

Solving:

`8C5=(8!)/(5!(8-5)!)=(8\cdot7\cdot6\cdot(5\cdot4\cdot3\cdot2\cdot1))/((5\cdot4\cdot3\cdot2\cdot1)\cdot(3)!)`

We can cancel the factors in the parentheses of numerator and denominator since they are the same:

`8C5=(8\cdot7\cdot6)/(3!)=(8\cdot7\cdot6)/(3\cdot2\cdot1)=(8\cdot7\cdot6)/(6)=8\cdot7=56`

The first combination is 56 then.

When tossing a coin 8 times, we have 56 different ways to obtain 5 tails.

Now we need to do the same process for 6 and 7 tails and add all 3 results at the end.

Calculating the number of possible ways to obtain 6 tails out of 8:

`8C6=(8!)/(6!\cdot2!)=(8\cdot7\cdot6!)/(6!\cdot2!)=(8\cdot7)/(2)=28`

When tossing a coin 8 times, we have 28 different ways to obtain 6 tails.

Now for 7 tiles:

`8C7=(8!)/(7!\cdot1!)=(8\cdot7!)/(7!)=8`

When tossing a coin 8 times, we have 8 different ways to obtain 7 tails.

Finally, the number of ways a person can toss a coin 8 times and obtain between 5 and 7 tails, inclusive is:

`56+28+8=92\text{ ways}`

Then, the answer is 92 ways.