7 candidates are to be examined 2 in mathematics and the remaining in different subjects . in how many ways can they be seated in a row so that the two examinees in mathematics may not sit together ?
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Solution:-
When there is no restriction in the sitting arrangements, the total number of ways in which 7 candidates can sit = 7p₇ ways or 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 ways.
When two candidates of mathematics sit together, we will consider them as one candidate. Now, the total number of candidates become 6. And they can be seated in 6p₆ ways or 6 × 5 × 4 × 3 × 2 × 1 = 720 ways but the two candidates of mathematics can be arranged among themselves in 2p₂ ways or 2 × 1 = 2 ways.
So, the total number of ways in which the two candidates of mathematics sit together = 720 × 2 = 1440 ways.
Hence, the number of ways in which the two mathematics candidates do not sit together = 5040 - 1440 = 3600 ways.
Answer.
When there is no restriction in the sitting arrangements, the total number of ways in which 7 candidates can sit = 7p₇ ways or 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 ways.
When two candidates of mathematics sit together, we will consider them as one candidate. Now, the total number of candidates become 6. And they can be seated in 6p₆ ways or 6 × 5 × 4 × 3 × 2 × 1 = 720 ways but the two candidates of mathematics can be arranged among themselves in 2p₂ ways or 2 × 1 = 2 ways.
So, the total number of ways in which the two candidates of mathematics sit together = 720 × 2 = 1440 ways.
Hence, the number of ways in which the two mathematics candidates do not sit together = 5040 - 1440 = 3600 ways.
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Answer:
above answer is correct
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