Question

Number of ways in which \[8\] people can be arranged in a line if \[A\] and \[B\] must be next each other and \[C\] must be somewhere behind \[D\] is equal to
A.\[10080\]
B.\[5040\]
C.\[5050\]
D.\[10100\]

Answer 1

Hint: A permutation is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. The formula for a permutation is \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;}}\]where, \[n\] is the total items in the set and \[r\]is the number of items taken for the permutation.
Here we have to assume the people who are sitting together as one. Since \[C\] must be somewhere behind \[D\]. Therefore, we need to consider all cases according to positions of \[D\]. Then find the number of ways the people sit together. Then the total number of ways is the number of ways the people sit together times the sum of the ways obtained in all possible cases.

Complete step-by-step answer:
Given \[A\] and \[B\] must be next to each other and \[C\] must be somewhere behind \[D\].
Since \[A\] and \[B\] are next to each other so we assume them as one. So now total people will be \[7\].
Now \[C\] should be somewhere behind \[D\], therefore we need to consider all cases according to positions of \[D\].
case(i): When \[D\] is at first position \[C\] can be placed at any remaining \[6\] positions=Total ways to arrange remaining \[6\] people (\[D\] is at first) \[ = 6!\].
case(ii): When \[D\] is at second position \[C\] can be placed only at remaining \[5\] positions after \[D\], so at first position there can be only \[5\] people\[ = \]Total ways will be\[ = {}^5{P_1} \times 5! = 5 \times 5!\].
case(iii): When \[D\] is at third position \[C\] can be placed only at remaining \[4\] positions after \[D\], so at first and second there can be only \[{}^5{P_2} = 5 \times 4\] choices\[ = \]Total ways will be\[ = 5 \times 4 \times 4!\]
Similarly,
case(iv): When \[D\] is at fourth position, Total ways \[ = {}^5{P_3} \times 3! = 5 \times 4 \times 3 \times 3!\]
case(v): When \[D\] is at fifth position, Total ways=\[ = {}^5{P_4} \times 2! = 5 \times 4 \times 3 \times 2 \times 2!\]
case(vi): When \[D\] is at sixth position, Total ways=5\[ = {}^5{P_5} \times 1! = 5 \times 4 \times 3 \times 2 \times 1 \times 1!\]
case(vii): \[D\] can't be at seventh position because that's the last position so then \[C\] can't be behind \[D\].
Now as \[A\] and \[B\] together can sit in two ways \[AB\] or \[BA\].
Hence, total number of ways\[ = 2 \times \left\{ {(i) + (ii) + (iii) + (iv) + (v) + (vi)} \right\}\]\[
   = 2 \times \left\{ {6! + 5 \times 5! + 5 \times 4 \times 4! + 5 \times 4 \times 3 \times 3! + 5 \times 4 \times 3 \times 2 \times 2! + 5 \times 4 \times 3 \times 2 \times 1 \times 1!} \right\} \\
   = 2 \times \left\{ {720 + 600 + 480 + 360 + 240 + 120} \right\} \\
   = 5040 \;
 \]
Hence, the Option (B) is correct.
So, the correct answer is “Option B”.

Note: Note that the factorial of \[n\] is denoted as \[n!\]. The factorial of a natural number is a number multiplied by "number minus one", then by "number minus two", and so on till \[1\] i.e., \[n! = n \times \left( {n - 1} \right) \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times - - - \times 2 \times 1\] .
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where we select the items in any order. Using the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;r!\;}}\]where, \[n\] is the total items in the set and \[r\]is the number of items taken for the permutation find the value of \[n\].