How many ways can a person toss a coin 8 times so that the number of heads is between 4 and 7 inclusive?

Answer 1

Let us break this question down into its components.

How many different ways can a person toss a coin 8 times and get 7 heads?

In other words, how many different sequences of 8 tosses are there that contain exactly 7 heads?

This is equivalent to asking "how many different sequences of 8 tosses contain exactly 1 tail?"

We can actually write all of them down:

THHHHHHH

HTHHHHHH

HHTHHHHH

HHHTHHHH

HHHHTHHH

HHHHHTHH

HHHHHHTH

HHHHHHHT

Those are 8 different sequences.

Surely there must be an easier way than writing all the possibilities... There is!

The logic is as follows: there are 8 available slots (due to the 8 tosses), the single tail has to go in 1 of the 8 slots and the remaining 7 slots are all heads. There are 8 choices for where the single tail can go so there are 8 different sequences.

How many different ways can a person toss a coin 8 times and get 6 heads?

In other words, how many different sequences of 8 tosses are there that contain exactly 6 heads?

This is equivalent to asking "how many different sequences of 8 tosses contain exactly 2 tails?"

There are 8 available slots, the 2 tails have to go in 2 of the 8 slots and the remaining slots are all heads.

How many different choices are there for the 2 slots (where the 2 tails will go) out of the 8 available slots?

The answer to that is precisely (8 choose 2).

(8 choose 2) = 8! / (2! (8-2)!) = 8! / (2! 6!) = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (2 x 1 x 6 x 5 x 4 x 3 x 2 x 1) = (8 x 7) / (2 x 1) = 56 / 2 = 28

We did not have to go through the tails. We can also think about this question as follows:

There are 8 available slots, the 6 heads have to go in 6 of the 8 slots and the remaining slots are all tails.

How many different choices are there for the 6 slots (where the 6 heads will go) out of the 8 available slots?

The answer to that is precisely (8 choose 6).

(8 choose 6) = 8! / (6! (8-6)!) = 8! / (6! 2!) = 8! / (2! 6!) = 28

It might be easier to think about where to put 2 objects as opposed to where to put 6 objects (in a line with 8 slots), but the tasks are equivalent!

How many different ways can a person toss a coin 8 times and get 5 heads?

There are 8 available slots, the 5 heads have to go in 5 of the 8 slots and the remaining slots are all tails.

How many different choices are there for the 5 slots (where the 5 heads will go) out of the 8 available slots?

The answer to that is precisely (8 choose 5).

(8 choose 5) = 8! / (5! (8-5)!) = 8! / (5! 3!) = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1 x 5 x 4 x 3 x 2 x 1) = (8 x 7 x 6) / (3 x 2 x 1) = 336 / 6 = 56

How many different ways can a person toss a coin 8 times and get 4 heads?

There are 8 available slots, the 4 heads have to go in 4 of the 8 slots and the remaining slots are all tails.

How many different choices are there for the 4 slots (where the 4 heads will go) out of the 8 available slots?

The answer to that is precisely (8 choose 4).

(8 choose 4) = 8! / (4! (8-4)!) = 8! / (4! 4!) = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (4 x 3 x 2 x 1 x 4 x 3 x 2 x 1) = (8 x 7 x 6 x 5) / (4 x 3 x 2 x 1) = 1680 / 24 = 70

The answer is thus:

8 + 28 + 56 + 70 = 162

In summary, we can obtain the answer with the following summation:

∑_k=4^k=7 (8 choose k) = (8 choose 4) + (8 choose 5) + (8 choose 6) + (8 choose 7) = 162.

Please let me know if you need any clarification.