Question

if there are 6 married couples and we have to select 4 people out of these 6 couples such that we get at least one couple. Find no of ways this can be done

• A. 255
• B. 240
• C. 495
• D. 210

Answer 1

Answer: (A)

255

Answer 2

To find the number of ways to select 4 people from 6 married couples such that at least one couple is selected, we can use the principle of complementary counting.

There are 6 married couples, which means there are a total of $6 \times 2 = 12$ people.

1. Calculate the total number of ways to select 4 people from the 12 people. This can be calculated using the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Total ways = $\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.

2. Calculate the number of ways to select 4 people such that NO couple is selected. For no couple to be selected, all 4 people must come from different couples.

   • First, choose 4 couples out of the 6 available couples: $\binom{6}{4}$ ways. $\binom{6}{4} = \binom{6}{2} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$
   • From each of these 4 chosen couples, select one person. Since each couple has 2 people, there are $\binom{2}{1}$ ways to select one person from each couple. As we have 4 couples, this is $\binom{2}{1} \times \binom{2}{1} \times \binom{2}{1} \times \binom{2}{1} = 2^4$. $\binom{2}{1}^4 = 2^4 = 16$
   • The total number of ways to select 4 people with no couple is the product of these two steps: Ways (no couple) = $\binom{6}{4} \times \binom{2}{1}^4 = 15 \times 16 = 240$.

3. Calculate the number of ways to select 4 people with at least one couple. This is the total number of ways minus the number of ways where no couple is selected. Ways (at least one couple) = Total ways - Ways (no couple) $\text{Ways (at least one couple)} = 495 - 240 = 255$