Combinatorial analysis

A CAR COMPORTA TWO PASSENGERS IN BANK OF FRONT AND BACK ON THREE BANK. CALCULATE THE NUMBER OF DIFFERENT ALTERNATIVES FOR THE AUTOMOBILE fill USING 7 PEOPLE, SO SUCH PEOPLE NEVER MIND A PLACE IN FRONT SEATS

THE PROBLEM IS RESOLVED AS FOLLOWS:

There are 7 people, and one can never go in the front seat.

Let's call this person of John, for example.

So let's first calculate the number of ways to fill the car WITHOUT John, using the other six only:

As we have 6 people and 5 people in the car then we calculate the array of 6 elements, taken 5-5:

A6,5 = 720

Now let's calculate the number of ways to fill the car WITH John.

We know that John may not be in the front seats, so it should be in one of three banks back.

Then we fix the John in one of the rear seats (4 places left over then in the car), and then calculate the number of ways to put the other 6 people in those four places, that is, an array of six elements, taken 4-4:

A6,4 = 360

The John may be in any of the three rear seats, so we should multiply that result by 3:

3 x A6,4 = 3 x 360 = 1080

The total number of ways to fill the car is the sum of the two arrangements (COM John and John SEM).

So the total number is 720 + 1080 = 1800 ways !!!