The ten students in a club are lined up in a row for a group photograph. How many different arrangements are possible if the club includes one set of identical triplets wearing matching clothes? 604,800 720 10,000,000

Respuesta :

It's 10!/3! so its 604, 800.

Answer:

604800

Step-by-step explanation:

Given : The ten students in a club are lined up in a row for a group photograph.

To Find:  How many different arrangements are possible if the club includes one set of identical triplets wearing matching clothes?

Solution:

The club includes one set of identical triplets wearing matching clothes

So, the remaining students = 10-3 = 7

Now there is an arrangement between these seven students.

Since order has to be maintained .So, we will use permutation

Formula : [tex]^nP_r=\frac{n!}{(n-r)!}[/tex]

So,  [tex]^{10}P_7=\frac{10!}{(10-7)!}[/tex]

[tex]^{10}P_7=\frac{10!}{(3)!}[/tex]

[tex]^{10}P_7= 604800[/tex]

Hence there are 604800 possible arrangements  if the club includes one set of identical triplets wearing matching clothes