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V125BC [204]
2 years ago
6

Permutations and, combinations: problem type 2.

Mathematics
1 answer:
USPshnik [31]2 years ago
5 0

Step-by-step explanation:

<u>Permutations</u>:

n<em>P</em><em> </em>r = n!/(n + r)!

<u>Combinations</u>:

<em>nCr</em> = n!/(n - r)!r!

c - combination

p - permutations

Since in both the problems they are looking for different groups, it is a combination problem. the order in which the people were selected in each group is not important, what is required are different groups.

<u>A.</u>

n = 51

r = 3

51<em>C</em><em> </em>3 = 51! / ((51 - 3) ×3!)

<em>51C</em><em>3</em><em> </em><em>=</em><em> </em><em>2</em><em>0</em><em> </em><em>8</em><em>2</em><em>5</em><em> </em>groups

<u>OR:</u>

c =  \frac{51}{3}  \times  \frac{50}{2}  \times 49

c = 20825

<em><u>B</u></em><em><u>.</u></em>

c =  \frac{14}{4}  \times  \frac{13}{3}  \times  \frac{12}{2}  \times 11

c = 1001

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