Alicia has 7 paintings she wants to hang side by side on her wall.
2. If Alicia only wants to display 4 of the paintings, in how many ways can she choose the paintings she wishes to display? Show your work.
3. In how many ways can Alicia arrange 4 out of the 7 paintings? Show your work.

The question is an illustration of permutation and combinations.

The given parameters are:

[tex]\mathbf{n = 7}[/tex]

[tex]\mathbf{r = 4}[/tex]

(a) Choose 4 of 7 paintings

Choosing 4 of 7 paintings implies combination.

So, we have:

[tex]\mathbf{^nC_r = \frac{n!}{(n - r)!r!}}[/tex]

Substitute values for n and r

[tex]\mathbf{^7C_4 = \frac{7!}{(7 - 4)!4!}}[/tex]

[tex]\mathbf{^7C_4 = \frac{7!}{3!4!}}[/tex]

Expand

[tex]\mathbf{^7C_4 = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4 !}}[/tex]

[tex]\mathbf{^7C_4 = 35}[/tex]

Hence, the number of ways she can choose 4 out of the 7 paintings is 35

(b) Arrange 4 of 7 paintings

Arranging 4 of 7 paintings implies permutation.

So, we have:

[tex]\mathbf{^nP_r = \frac{n!}{(n - r)!}}[/tex]

Substitute values for n and r

[tex]\mathbf{^7P_4 = \frac{7!}{(7 - 4)!}}[/tex]

[tex]\mathbf{^7P_4 = \frac{7!}{3!}}[/tex]

Expand

[tex]\mathbf{^7P_4 = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}}[/tex]

[tex]\mathbf{^7P_4 = 840}[/tex]

Hence, the number of ways she can arrange 4 out of the 7 paintings is 840

Read more about permutation and combinations at:

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