Question

The number of ways in which $ 6 $ men can be arranged in arrow so that three particular men are consecutive, is

Answer 1

Hint: As we can see in the question we have to find the number of ways by arranging them, so it means we have to apply permutations. The formula of permutation of $ n $ things taken $ r $ at a time is $ ^n{P_r} = \dfrac{{n!}}{{(n - r)!}} $ . We will first arrange them according to the question and then we will assume that three particular men who are arranged to be consecutive as one.

Complete step by step solution:
As per the question we have total number of men
  $ 6 $
We have to arrange them in such a way that three particular men are to be consecutive.
So let us consider those $ 3 $ men as a single item.
Now when we take those $ 3 $ men out of $ 6 $ , we are left with total
  $ 6 - 3 = 3 $ men.
Also we have considered those three men as one, i.e. $ 1 $ , So by including this one item and other three, we have total of
 $ 3 + 1 = 4 $ items.
Now those $ 4 $ items can be arranged in total four ways, i.e.
 $ 4! $ or $ ^4{P_4} $ .
We can solve it by applying the formula of permutation, by comparing we have
 $ n = 4,r = 4 $
It can be solved as
 $ \dfrac{{4!}}{{(4! - 4!)}} $ .
On solving we have:
\[\dfrac{{4 \times 3 \times 2 \times 1}}{{0!}}\].
It gives the value
 $ 24 $ ways.
And those $ 3 $ men which are considered as single, can be arranged among themselves in three ways:
 $ 3! $ .
It can also be written as $ ^3{P_3} $ .
Again we can solve it by applying the formula of permutation, by comparing we have
 $ n = 3,r = 3 $
It can be solved as
 $ \dfrac{{3!}}{{(3! - 3!)}} $ .
On solving we have:
\[\dfrac{{3 \times 2 \times 1}}{{0!}} = 6\].
It gives the value
 $ 6 $ ways.
So total number of ways in which they can be arranged are
 $ ^4{P_4}{ \times ^3}{P_3} $ .
We can put the value of both from the above i.e.
 $ 24 \times 6 = 144 $ .
Hence the required answer is $ 144 $ ways.
So, the correct answer is “$ 144 $ ways”.

Note: We should note that the value of $ 0! $ is one, i.e. $ 1 $ . Whenever we get this type of question, we should carefully read the question and identify what should be applied permutation or combination. Combination is a way of selecting items from a collection where the order of selection does not matter. The formula of combination is $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ . In permutation the order does not matter but in combination it does matter.