Answer
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Hint: The no. of ways in which 6 persons can stand in a queue is same as the number of arrangements of 6 different things taken all at a time, which is ordered pair is, do permutation, \[{}^n{P_r}\]
Complete step-by-step answer:
We are given a total of 6 persons standing in a queue. We need to find the no. of ways in which we can arrange these 6 persons to stand in the queue.
The no. of ways in which 6 persons can stand in a queue is the same as the no. of arrangements of 6 different things taken all the time. Now, this can be done through Permutation.
Now, the required no. of ways = \[{}^6{P_6}\]
Where, n = 6 and r = 6.
Thus, applying the formula, we get
\[\begin{array}{l}^n{P_r}{\rm{ = }}{}^6{P_6}{\rm{ = }}\dfrac{{6!}}{{(6 - 6)!}}{\rm{ = }}\dfrac{{6!}}{{0!}}{\rm{ = }}\dfrac{{6!}}{1}\\\\{}^6{P_6}{\rm{ = 6! = 6 x 5 x 4 x 3 x 2 x 1}}\\\\{\rm{ = 720}}\\\end{array}\]
Thus, there are 720 ways in which 6 persons can stand in a queue.
Note: You may confuse this question of permutation with that of ambition. If you do decombination, you get the answer as,
\[\begin{array}{l}{}^n{C_r}{\rm{ = }}\dfrac{{n!}}{{(n - r)!{\rm{ r!}}}}\\{}^6{C_6}{\rm{ = }}\dfrac{{6!}}{{(6 - 6)!{\rm{ 6!}}}}{\rm{ = }}\dfrac{{6!}}{{0!{\rm{ 6!}}}}{\rm{ = 1}}\end{array}\]
We get no. of arrangements as 1, which is not possible. Thus, use permutation.
Complete step-by-step answer:
We are given a total of 6 persons standing in a queue. We need to find the no. of ways in which we can arrange these 6 persons to stand in the queue.
The no. of ways in which 6 persons can stand in a queue is the same as the no. of arrangements of 6 different things taken all the time. Now, this can be done through Permutation.
Now, the required no. of ways = \[{}^6{P_6}\]
Where, n = 6 and r = 6.
Thus, applying the formula, we get
\[\begin{array}{l}^n{P_r}{\rm{ = }}{}^6{P_6}{\rm{ = }}\dfrac{{6!}}{{(6 - 6)!}}{\rm{ = }}\dfrac{{6!}}{{0!}}{\rm{ = }}\dfrac{{6!}}{1}\\\\{}^6{P_6}{\rm{ = 6! = 6 x 5 x 4 x 3 x 2 x 1}}\\\\{\rm{ = 720}}\\\end{array}\]
Thus, there are 720 ways in which 6 persons can stand in a queue.
Note: You may confuse this question of permutation with that of ambition. If you do decombination, you get the answer as,
\[\begin{array}{l}{}^n{C_r}{\rm{ = }}\dfrac{{n!}}{{(n - r)!{\rm{ r!}}}}\\{}^6{C_6}{\rm{ = }}\dfrac{{6!}}{{(6 - 6)!{\rm{ 6!}}}}{\rm{ = }}\dfrac{{6!}}{{0!{\rm{ 6!}}}}{\rm{ = 1}}\end{array}\]
We get no. of arrangements as 1, which is not possible. Thus, use permutation.
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