Answer

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**Hint:**Based on this problem we have a combination formula in which we arrange (or) by selecting the objects appropriately.

Here we will use the combination formula in this problem.

Formula: \[n!\, = \,n\, \times \,\left( {n\, - \,1} \right)\, \times \,\left( {n\, - \,2} \right)\, \times \,...\, \times \,3\, \times \,2\, \times \,1\]

\[^n{C_{r\,}} = \,\dfrac{{n!}}{{\left( {n - r} \right)!r!}}\,\,\,;\,\,0 \leqslant r < n\] And

\[^n\,{C_{o\,}} = n{C_n}\, = 1\]

Also, \[\,^n\,{C_1}\, = \,n\]

**Complete step by step answer:**

Based on the given problem,

Given that: \[4\,\] pants, \[3\] shirts and \[2\] banians

So, probably a man can put them in some number of ways.

That is,

\[\left( {^4\,{C_1} \times\,^3{C_1}\, \times \,^2{C_1}} \right)\] ways

Now, expanding the above expression using the formula \[^n{C_1}\, = \,n\] , we get \[^4{C_1} = 4,\,\,^3{C_1} = 3\,\] and \[^2{C_1}\, = 2\]

And so we have,

\[\left( {4\, \times \,3\, \times \,2} \right)\] ways

On simplifying the above expression by multiplying the each term we get

(24) ways

So, therefore we can conclude that the number of ways which he can put them on is \[\,24\] ways.

**Note:**The number of clothes in which he can put on him is \[\,24\] ways, Let us discuss the above problem as recap (or) review it at glance. Here in this problem, if he has \[4\] pants, then he has to select all the four pants which he has to wear on him. That is, he can select the \[4\] pants in \[4\] ways, similarly, if he has \[3\] shirts, then he instantly pick/select the \[3\] shirts in \[3\] ways, also if he has \[2\] baniyans and has to select it then he can do it in \[2\] ways. Thus, in general if he picks \[a,b,\] and \[c\] objects all one at a time then, the total number of ways will be chosen as \[\left( {a\, \times \,b\, \times \,c} \right)\] ways. And similarly here we arrive at the conclusion as the total number of ways he can put the cloth on him is \[\left( {4\, \times \,3\, \times \,2} \right)\] ways which is equals to \[24\] ways.

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