Find the number of ways of getting 2 heads and 4 tails in 6 throws of a coin.

Answer

To find the number of ways of getting 2 heads and 4 tails in 6 throws of a coin, we can use the concept of permutations and combinations. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to find the number of distinct arrangements of 2 heads (H) and 4 tails (T) in a sequence of 6 coin tosses.

### Step 2: Total Outcomes
The total number of outcomes (arrangements) can be calculated using the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{n!}{n_1! \times n_2!} \]
where:
- \(n\) is the total number of items,
- \(n_1\) is the number of identical items of one type (heads),
- \(n_2\) is the number of identical items of another type (tails).

### Step 3: Substitute the Values
In our case:
- \(n = 6\) (total number of coin tosses),
- \(n_1 = 2\) (number of heads),
- \(n_2 = 4\) (number of tails).

So, we can write:
\[ \text{Number of arrangements} = \frac{6!}{2! \times 4!} \]

### Step 4: Calculate Factorials
Now, we calculate the factorials:
- \(6! = 720\)
- \(2! = 2\)
- \(4! = 24\)

### Step 5: Substitute Factorials into the Formula
Now substitute these values back into the formula:
\[ \text{Number of arrangements} = \frac{720}{2 \times 24} \]

### Step 6: Simplify the Expression
Calculating the denominator:
\[ 2 \times 24 = 48 \]
Now, substitute this into the equation:
\[ \text{Number of arrangements} = \frac{720}{48} \]

### Step 7: Perform the Division
Now, perform the division:
\[ \frac{720}{48} = 15 \]

### Conclusion
Thus, the number of ways to get 2 heads and 4 tails in 6 throws of a coin is **15**.

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