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In how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include atleast two teachers?
- a)220
- b)672
- c)896
- d)968
Correct answer is option 'B'. Can you explain this answer?
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In how many ways can a selection of 6 out of 4 teachers and 8 students...
Problem: In how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include at least two teachers?
Solution:
To solve this problem, we need to consider two cases:
Case 1: Exactly 2 teachers are selected
In this case, we need to select 2 teachers from 4 and 4 students from 8. This can be done in:
${4 \choose 2} \times {8 \choose 4} = 6 \times 70 = 420$ ways
Case 2: Exactly 3 teachers are selected
In this case, we need to select 3 teachers from 4 and 3 students from 8. This can be done in:
${4 \choose 3} \times {8 \choose 3} = 4 \times 56 = 224$ ways
Total number of ways to select at least 2 teachers:
$420 + 224 = 644$
But we need to select a total of 6 people, so we can also select 4 teachers and 2 students or 5 teachers and 1 student. These cases can be done in:
${4 \choose 4} \times {8 \choose 2} = 28$ ways (4 teachers and 2 students)
${4 \choose 5} \times {8 \choose 1} = 32$ ways (5 teachers and 1 student)
Therefore, the total number of ways to select 6 people with at least 2 teachers is:
$644 + 28 + 32 = 672$
Hence, the correct option is (b) 672.
Solution:
To solve this problem, we need to consider two cases:
Case 1: Exactly 2 teachers are selected
In this case, we need to select 2 teachers from 4 and 4 students from 8. This can be done in:
${4 \choose 2} \times {8 \choose 4} = 6 \times 70 = 420$ ways
Case 2: Exactly 3 teachers are selected
In this case, we need to select 3 teachers from 4 and 3 students from 8. This can be done in:
${4 \choose 3} \times {8 \choose 3} = 4 \times 56 = 224$ ways
Total number of ways to select at least 2 teachers:
$420 + 224 = 644$
But we need to select a total of 6 people, so we can also select 4 teachers and 2 students or 5 teachers and 1 student. These cases can be done in:
${4 \choose 4} \times {8 \choose 2} = 28$ ways (4 teachers and 2 students)
${4 \choose 5} \times {8 \choose 1} = 32$ ways (5 teachers and 1 student)
Therefore, the total number of ways to select 6 people with at least 2 teachers is:
$644 + 28 + 32 = 672$
Hence, the correct option is (b) 672.
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