Answer:
The number of different combinations of three students that are possible is 35.
Step-by-step explanation:
Given that three out of seven students in the cafeteria line are chosen to answer a survey question.
The number of different combinations of three students that are possible is given as:
7C3 (read as 7 Combination 3)
xCy (x Combination y) is defines as
x!/(x-y)!y!
Where x! is read as x - factorial or factorial-x, and is defined as
x(x-1)(x-2)(x-3)...2×1.
Now,
7C3 = 7!/(7 - 3)!3!
= 7!/4!3!
= (7×6×5×4×3×2×1)/(4×3×2×1)(3×2×1)
= (7×6×5)/(3×2×1)
= 7×5
= 35
Therefore, the number of different combinations of three students that are possible is 35.
There are 35 different combinations to chose 3 students from the 7 students
The given parameters are:
Students, n = 7
Selected , r = 3
The number of different combinations is calculated using:
Combination = nCr
This gives
Combination = 7C3
Using the combination formula, we have:
Combination = 35
Hence, there are 35 different combinations to chose 3 students from the 7 students
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