MHT CET · Maths · Permutation Combination
The number of ways in which 5 boys and 3 girls can be seated on a round table, if a particular boy \(B_1\) and a particular girl \(G_1\) never sit adjacent to each other, is
- A 7 !
- B \(5 \times 6\) !
- C \(6 \times 6\) !
- D \(5 \times 7\) !
Answer & Solution
Correct Answer
(B) \(5 \times 6\) !
Step-by-step Solution
Detailed explanation
First, we arrange 4 boys and 2 girls (excluding \(\text B _1\) and \(\text G _1\) ) around the table, which can be done in \(5!\) ways.
In any such arrangement, \(\text B _1\) and \(\text G _1\) can be arranged in 6 available gaps in \({ }^6 P _2=6 \times 5\) ways.
\(\therefore\) Total number of arrangements \(=5!\times 6 \times 5\)
\(=6!\times 5\)
In any such arrangement, \(\text B _1\) and \(\text G _1\) can be arranged in 6 available gaps in \({ }^6 P _2=6 \times 5\) ways.
\(\therefore\) Total number of arrangements \(=5!\times 6 \times 5\)
\(=6!\times 5\)
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