The number of students are 11.
We know that it is the number of ways of grouping of 4 persons out of 11 persons.Choosing or grouping is different from arranging.
We can arange 4 persone from 11 on 4 chairs 11P4 ways = 11*10*9*8 ways.The reason is that the number of choices of arranging on the 1st chair is 11 and the number of choices of 2nd person from the remaining (11-1) = 10 persons is 10, and in this way for the 3rd and 4 th chair the number of choices are 9 and 8.
Now let us pressume that in all possible ways it is x number of different groups in which we can select 4 persons in each group from 11 persons. In any of these group of 4 persons could be arranged in 4P4 = 4!. Therefore the total possible number of ways we can arrange 4 persons from 11 persons = x * 4!.
Therefore 11P4 = x!*4!
Therefore k = 11P4/4! = 11*10*9*8/(4*3*2*1 )= 330 ways.
Therefore the different ways of making 4 person group to transport is 330 ways.
But the question is pertaining to the number of ways of grouping 4 persons out of 7 remaining persons, as in the 1st trip 4 persons have already been transported.
So applying the same rule , we get :
x = 7P4/4! = 7*6*5*4 /(4*3*2*1) = 35 ways.
In the given problem, the Taxi has to shuttle 11 students to a concert and it can only hold 4 students at a time. After the first trip, 4 students have been dropped off and now 11 - 4 = 7 remain.
Now the taxi driver has four seats and 7 students to choose from.
So for the first seat he has 7 options.
For the second seat he has 6 options.
For the third seat he has 5 options.
And for the fourth seat he has 4 options.
Therefore totally he has 7 * 6 * 5 * 4 = 840 options.
But in a taxi does it make a difference which seat you are sitting in? If it doesn't, the total number of ways decreases and becomes: 840/ (4*3*2*1) = 35.
It terms of combinations it is 7C4 = 7!/ 4! = 35.
Therefore the four students for the second trip can be chosen in 35 ways.