Question

There are 8 books on a shelf. If 3 books are chosen at random, how many different groups of3 books could be chosen? Determine if it is permutation or combination then solve.B). 28A). 19C). 56D). 65

Answer 1

The number of different groups of 3 books that can be chosen from 8 books is 56, calculated using the combination formula C(n, k) = n! / (k!(n - k)!), resulting in C(8, 3) = 56.

Step-by-step explanation:

The question is asking to find out how many different groups of 3 books can be chosen from 8 books on a shelf. This is a combinations problem, not a permutation problem, because the order of selection does not matter. To solve this problem, we use the combination formula, which is:

C(n, k) = n! / (k!(n - k)!)

Where n is the total number of items to choose from, in this case, 8 books, and k is the number of items to choose, which is 3 books. So:

C(8, 3) = 8! / (3!(8 - 3)!) = (8 × 7 × 6) / (3 × 2 × 1) = 56.

Therefore, there are 56 different groups of 3 books that could be chosen, which corresponds to option C).

Answer 2

Step 1:

The permutation is the number of different arrangement which can be made by picking r number of things from the available n things. The combination is the number of different groups of r objects each, which can be formed from the available n objects.

Permutation represent arrangement

Combination represent selection

Step 2:

The question is combination because its involved chosen or selection.

Step 3:

Number of ways of selecting 3 books from 8 books

n = 8 and r = 3

`\begin{gathered} ^{}^{}Numberofwaysofselectingrobjectsoutofnobjects=^nC_r \\ =\text{ }(n!)/((n-r)!r!) \\ =\text{ }(8!)/((8-3)!3!) \\ =\text{ }(8!)/(5!3!) \\ =\text{ }(8*7*6*5!)/(5!*3*2*1) \\ \text{Cancel out common factors} \\ =\text{ }(8*7*6)/(6) \\ =\text{ 56 ways} \end{gathered}`

Final answer

Option C 56