Suppose we want to choose 2 objects, without replacement, from the 5 objects pencil, eraser, desk, chair, and lamp. (a)How many

Question

2 months ago

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Suppose we want to choose 2 objects, without replacement, from the 5 objects pencil, eraser, desk, chair, and lamp. (a)How many

ways can this be done, if the order of the choices matters? (b)How many ways can this be done, if the order of the choices does not matter?

Mathematics

2 answers:

Leona [12.6K] · Answer 1 · 2026-03-10T06:20:47+03:00

Leona [12.6K]2 months ago

3 0

Answer:

20

10

Step-by-step explanation:

I just completed the test

Inessa [12.5K] · Answer 2 · 2026-03-10T06:22:55+03:00

Answer:

a) 20 different arrangements

b) 10 different arrangements

Step-by-step explanation:

When the order of selection is significant, we utilize permutations to determine the number of arrangements; if the order does not matter, we apply combinations.

For part a)

We must select 2 items from a collection of 5, considering that order matters. This is a permutation scenario, thus we calculate 5P2.

The general formula for permutations of n items taken r at a time is:

$nPr=\frac{n!}{(n-r)!}$

By inserting n=5 and r=2, we have:

$5P2=\frac{5!}{(5-2)!} =20$

Hence, it is possible to select 2 items from a group of 5 when the selection order is relevant.

For part b)

Since the order of selection is unimportant, we use combinations to calculate the number of ways of selecting 2 items from 5, represented as 5C2.

The general formula for combinations of n items taken r at a time is:

$nCr=\frac{n!}{r!(n-r)!}$

Plugging n=5 and r=2 into the formula results in:

$5C2=\frac{5!}{2!(5-2)!} =10$

This demonstrates that we can choose 2 items from a collection of 5 when the order of selection is not significant.