Question

If 12 schools teams are participating in a quiz contest,then the number of ways the first,second and third positions may be won is

Answer 1

Answer:

In this there are 12 teams from which only 3 are selected as there are only 3 positions i.e.first,second and third

Therefore, 12p3=12!/(12-3)!=12!/9!

= 12×11×10×9!/9!= 12×11×10=1320

Answer 2

The total number of ways are 1320.

Step-by-step explanation:

Consider the provided information.

The total number of schools are 12.

The order of schools matter therefore we will use the concept of permutation.

$^nP_r=\dfrac{n!}{(n-r)!}$

Here we have 3 positions for 12 schools.

Substitute n = 12 r=3 in above formula.

$^{12}P_3=\dfrac{12!}{(12-3)!}$

$^{12}P_3=\dfrac{12!}{9!}$

$^{12}P_3=\dfrac{9!\times10\times11\times12}{9!}$

$^{12}P_3=10\times11\times12$

$^{12}P_3=1320$

Hence, the total number of ways are 1320.

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