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**Hint:**From the given data it is clear that the first row consists of 6 girls so these 6 girls can be arranged in $6!$ possible ways. And in the second row there are 20 boys with the corner spots fixed for the tallest boys of the class so they can be arranged for $2!$ ways and the remaining 18 boys can be arranged in $18!$ ways. So, the total ways in which the students can be arranged will be the product of these values.

**Complete step-by-step solution:**

In this question, we are given that a class photograph is to be taken with the front row consisting of 6 girls and 20 boys standing behind them and the corner two positions are reserved for the two tallest boys of the class. We need to find out how many ways these students can be arranged.

So, according to the given data it is clear that the first row consists of 6 girls.

So, let us find ways for the first row of girls to be arranged.

$ \Rightarrow $Number of ways for 6 girls $ = 6!$

Now, the second row consists of 20 boys standing but the corner spots are fixed for the two tallest boys of the class.

$ \Rightarrow $Number of these 2 boys can be arranged$ = 2!$

So, the remaining 18 boys will be arranged in

$ \Rightarrow $Number of ways for 18 boys $ = 18!$

So, the total number of ways all these students can be arranged is given by

$ \Rightarrow $Total number of possible arrangements $ = 6! \times 2! \times 18! = 18! \times 1440$

**Hence, the students can be arranged for $18! \times 1440$ ways.**

**Note:**Here, note that

$ \Rightarrow n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 1$

So, therefore, 18! Will be equal to

$ \Rightarrow 18! = 18 \times 17 \times 16 \times ...... \times 1$

Here for arrangement we could also use the formula of permutations as ${}^nP_r = \dfrac{n!}{{n-r}!}$ but sometimes students are confused b/w permutation and combination so for solving these types of problems we should know all the formulae.

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