Q: A coin is tossed 9 times. How many ways could the result be 6 heads and 3 tails?
Answer is 84 ways
If you could explain the process on how you get the answer that would be really helpful because I always get stuck on this kind of question... thanks!
C(9,6) = C(9,3) = 84
or combinations of 9 items taken 6 (or 3) at a time
you have 9 options, 6 are the same and 3 are the same, so you do 9!/6!3!=84
Thank you @jollyrancher I don't know why I didn't do that before...
welcome
A: Ah, the coin toss conundrum! Well, let's see if I can put a smile on your face while explaining the process.
To find the number of ways to get 6 heads and 3 tails, we can use a little mathematical trickery. Since there are two possible outcomes (heads or tails) for each of the 9 coin tosses, we can think of it as a binomial expansion.
In this case, we're interested in the coefficient of the term with a power of 6 for the expansion of (H + T)^9, where H represents getting a head and T represents getting a tail. The binomial expansion formula is:
(n choose k) * (x^k) * (y^(n-k))
Where n is the total number of coin tosses (9 in this case) and k is the number of heads we want (6). The coefficient (n choose k) gives us the number of ways to choose k items from a set of n items.
Plugging in the numbers, we get:
(9 choose 6) * (H^6) * (T^3)
Now let's calculate the coefficient:
(9 choose 6) = 9! / (6! * (9-6)!) = 84
So, there are 84 ways to get 6 heads and 3 tails in 9 coin tosses! Don't forget to have some fun while counting those heads and tails though, because life's too short to be serious all the time.
To find the number of ways the result could be 6 heads and 3 tails when a coin is tossed 9 times, you can use the concept of combinations.
When tossing the coin 9 times, each toss has two possible outcomes - either a head or a tail. So, we have a total of 2 options for each toss.
Now, we need to determine the number of ways to choose 6 out of the 9 tosses to be heads, while the remaining 3 tosses will be tails. This can be solved using combinations.
The formula for combinations is given by:
C(n, r) = n! / (r! * (n-r)!)
Where n represents the total number of items, and r represents the number of items to be chosen.
In this case, we have n = 9 tosses, and we want to choose r = 6 heads. So the formula becomes:
C(9, 6) = 9! / (6! * (9-6)!)
Simplifying further:
C(9, 6) = 9! / (6! * 3!)
Using the factorial notation, where n! represents the product of all positive integers less than or equal to n:
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
3! = 3 * 2 * 1
We can cancel out some terms:
C(9, 6) = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / [(6 * 5 * 4 * 3 * 2 * 1) * (3 * 2 * 1)]
Cancelling out similar terms gives:
C(9, 6) = (9 * 8 * 7) / (3 * 2 * 1)
Simplifying further:
C(9, 6) = 504 / 6
C(9, 6) = 84
Therefore, there are 84 ways the result could be 6 heads and 3 tails when a coin is tossed 9 times.