A school's track and field team has 14 track athletes, and 10 field athletes. Due to budget constraints, the school is only allowed to send 3 athletes to the meet. how many ways can this be done if the coach wishes to send at least one track athlete. [3 marks/A]

1924 ways

Step-by-step explanation:

Let's calculate the three cases: one track athlete, two track athletes and three track athletes.

One track athlete:

The track athlete has 14 possible choices.

The two field athletes have a combination of 10 choose 2 possibilities:

C(10,2) = 10! / (8! * 2!) = 10*9/2 = 45

So we have 14 * 45 = 630 ways.

Two track athletes:

The two track athletes have a combination of 14 choose 2 possibilities:

C(14,2) = 14! / (12! * 2!) = 14*13/2 = 91

The field athlete has 10 possible choices.

So we have 91 * 10 = 930 ways.

Three track athletes:

The three track athletes have a combination of 14 choose 3 possibilities:

C(14,3) = 14! / (11! * 3!) = 14*13*12/6 = 364

So we have 364 ways.

Finally, the total number of ways is:

630 + 930 + 364 = 1924 ways.

here the number -5 is all of the following except a rational number.

step-by-step explanation:

i used to know how to do this in 7th but i forget easily

step-by-step explanation: