A taxi is shuttling 11 students to a concert. The taxi can hold 4 students.In how many ways can 4 students be chosen for the taxi's second trip

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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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.

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william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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