Question

A balanced coin with one side heads (H) and the other side tails (7) is repeatedly flipped, and the results are recorded. The coefficients in the expansion of (H+T)" give the number of ways to

get a particular combination of heads and tails torn flips of the coin. For example, in the expansion of (H+T), the term 21H³T means there are 21 ways to get exactly 2 heads and 5 tails
when flipping a coin 7 times. How many ways are there to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times?

Answer 1

There are 15 different ways to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times, as calculated by the binomial coefficient formula.

Step-by-step explanation:

The question asks how many ways there are to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times. This is a problem of combinatorics, specifically of calculating a binomial coefficient. The general formula for the binomial coefficient is given by C(n, k) = n! / (k!(n-k)!), where n is the total number of flips and k is the number of heads.

To find the answer to our question, we will use the formula with n = 6 (since the coin is flipped 6 times) and k = 2 (since we are looking for the number fo ways to get 2 heads). So, we calculate C(6, 2) = 6! / (2!(6-2)!) = 720 / (2! * 4!) = 720 / (2 * 24) = 720 / 48 = 15. Therefore, there are 15 different ways to get 2 heads and 4 tails in 6 flips of a balanced coin.

Answer 2

There are 15 ways to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times. This is calculated using the binomial coefficient C(6, 2), which equals 15.

Step-by-step explanation:

To find the number of ways to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times, we can use the binomial theorem which states that the coefficient in the expansion of (H+T)^n gives the number of ways to get a particular combination of heads and tails from n flips of the coin. The expansion of (H+T)^6 includes the term H^2T^4, and the coefficient of this term is the binomial coefficient C(6, 2), which is calculated using the formula:

C(n, k) = n! / (k!(n-k)!)

where n! denotes n factorial, and k is the number of successes (in this case, 2 heads).

So, C(6, 2) = 6! / (2!(6-2)!) = (6*5*4*3*2*1) / ((2*1)*(4*3*2*1)) = 720 / (2*24) = 720 / 48 = 15.

Therefore, there are 15 ways to get exactly 2 heads and 4 tails when flipping a coin 6 times.