Question
Find the number of ways to choose 5 chips from a bag containing 6 yellow chips and 8 red chips without replacement.
Answers
GPT-4o mini
To find the number of ways to choose 5 chips from a bag containing 6 yellow chips and 8 red chips, we start by defining the variables. Let \( y \) be the number of yellow chips chosen and \( r \) be the number of red chips chosen. We must satisfy the equation
\[
y + r = 5
\]
with the constraints:
\[
0 \leq y \leq 6 \quad \text{and} \quad 0 \leq r \leq 8.
\]
Since \( r = 5 - y \), we can rewrite the constraints as follows:
\[
0 \leq 5 - y \leq 8 \implies -5 \leq -y \leq 3 \implies 2 \leq y \leq 5.
\]
Thus, \( y \) can take the values 0, 1, 2, 3, 4, 5, or 6, but given our derived constraints, it can only be 2, 3, 4, or 5. We can now evaluate the number of combinations for each valid value of \( y \).
1. **When \( y = 2 \):**
\[
r = 5 - 2 = 3
\]
The number of ways to choose 2 yellow and 3 red chips is computed as:
\[
\binom{6}{2} \times \binom{8}{3}
\]
Calculate \( \binom{6}{2} \):
\[
\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15
\]
Calculate \( \binom{8}{3} \):
\[
\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
\]
Thus, the total for \( y = 2 \) is:
\[
15 \times 56 = 840.
\]
2. **When \( y = 3 \):**
\[
r = 5 - 3 = 2
\]
The number of ways to choose 3 yellow and 2 red chips is:
\[
\binom{6}{3} \times \binom{8}{2}
\]
Calculate \( \binom{6}{3} \):
\[
\binom{6}{3} = 20
\]
Calculate \( \binom{8}{2} \):
\[
\binom{8}{2} = 28
\]
Thus, the total for \( y = 3 \) is:
\[
20 \times 28 = 560.
\]
3. **When \( y = 4 \):**
\[
r = 5 - 4 = 1
\]
The number of ways to choose 4 yellow and 1 red chip is:
\[
\binom{6}{4} \times \binom{8}{1}
\]
Calculate \( \binom{6}{4} \):
\[
\binom{6}{4} = 15
\]
Calculate \( \binom{8}{1} \):
\[
\binom{8}{1} = 8
\]
Thus, the total for \( y = 4 \) is:
\[
15 \times 8 = 120.
\]
4. **When \( y = 5 \):**
\[
r = 5 - 5 = 0
\]
The number of ways to choose 5 yellow and 0 red chips is:
\[
\binom{6}{5} \times \binom{8}{0}
\]
Calculate \( \binom{6}{5} \):
\[
\binom{6}{5} = 6
\]
Calculate \( \binom{8}{0} \):
\[
\binom{8}{0} = 1
\]
Thus, the total for \( y = 5 \) is:
\[
6 \times 1 = 6.
\]
Now, we sum the results from all cases:
\[
840 + 560 + 120 + 6 = 1526.
\]
Therefore, the total number of ways to choose 5 chips from the bag is:
\[
\boxed{1526}.
\]
\[
y + r = 5
\]
with the constraints:
\[
0 \leq y \leq 6 \quad \text{and} \quad 0 \leq r \leq 8.
\]
Since \( r = 5 - y \), we can rewrite the constraints as follows:
\[
0 \leq 5 - y \leq 8 \implies -5 \leq -y \leq 3 \implies 2 \leq y \leq 5.
\]
Thus, \( y \) can take the values 0, 1, 2, 3, 4, 5, or 6, but given our derived constraints, it can only be 2, 3, 4, or 5. We can now evaluate the number of combinations for each valid value of \( y \).
1. **When \( y = 2 \):**
\[
r = 5 - 2 = 3
\]
The number of ways to choose 2 yellow and 3 red chips is computed as:
\[
\binom{6}{2} \times \binom{8}{3}
\]
Calculate \( \binom{6}{2} \):
\[
\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15
\]
Calculate \( \binom{8}{3} \):
\[
\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
\]
Thus, the total for \( y = 2 \) is:
\[
15 \times 56 = 840.
\]
2. **When \( y = 3 \):**
\[
r = 5 - 3 = 2
\]
The number of ways to choose 3 yellow and 2 red chips is:
\[
\binom{6}{3} \times \binom{8}{2}
\]
Calculate \( \binom{6}{3} \):
\[
\binom{6}{3} = 20
\]
Calculate \( \binom{8}{2} \):
\[
\binom{8}{2} = 28
\]
Thus, the total for \( y = 3 \) is:
\[
20 \times 28 = 560.
\]
3. **When \( y = 4 \):**
\[
r = 5 - 4 = 1
\]
The number of ways to choose 4 yellow and 1 red chip is:
\[
\binom{6}{4} \times \binom{8}{1}
\]
Calculate \( \binom{6}{4} \):
\[
\binom{6}{4} = 15
\]
Calculate \( \binom{8}{1} \):
\[
\binom{8}{1} = 8
\]
Thus, the total for \( y = 4 \) is:
\[
15 \times 8 = 120.
\]
4. **When \( y = 5 \):**
\[
r = 5 - 5 = 0
\]
The number of ways to choose 5 yellow and 0 red chips is:
\[
\binom{6}{5} \times \binom{8}{0}
\]
Calculate \( \binom{6}{5} \):
\[
\binom{6}{5} = 6
\]
Calculate \( \binom{8}{0} \):
\[
\binom{8}{0} = 1
\]
Thus, the total for \( y = 5 \) is:
\[
6 \times 1 = 6.
\]
Now, we sum the results from all cases:
\[
840 + 560 + 120 + 6 = 1526.
\]
Therefore, the total number of ways to choose 5 chips from the bag is:
\[
\boxed{1526}.
\]