Match the polynomial in the left column with its descriptive feature in the

right column.

A. x3 + 3x2 - 2x + 7

B. 3a b6

C. 3x4 - 9x3 + 5x8

D. 7a3b2 + 18ab2c – 9a3

E. 2x5 – 9x3 + 8x?

F. 4x8 – 7x2 + 9

G. x2 – 7

1. 9th degree monomial

II. Constant term of -7

III. 7th degree polynomial

IV. Leading coefficient of 4

V. Four terms

VI. 5th degree polynomial

VII. Equivalent to 5x8 + 3x4 – 9x3

(See explanation for further details)

Step-by-step explanation:

A. : V. Four terms.

B. : I . 9th Degree Monomial (Instead, 7th Degree Monomial)

C. : VII - Equivalent to

D. : III - 7th degree polynomial.

E. : VI - 5th degree polynomial.

F. : IV - Leading coefficient of 4.

G. : II - Constant term of -7.

First, if we have a set of K elements, such that are ordered as:

{x₁, x₂, ...}

The total number of permutations for the K elements can be found in the next way.

For the first element in the set, we have K options.

For the second element in the set, we have (K - 1) options (because we already choose one)

For the third element we have (K - 2) options, and so on.

The total number of permutation is equal to the product between the numbers of options for each position's element, then the number of permutations for K elements is:

permutations = K*(K - 1)*(K - 2)*....*2*1 = K!

Now suppose that we have a set of N elements, and we want to make groups of K elements.

The total number of different combinations of K elements is given by the equation:

In this case we have 15 objects (then  N = 15) and we take 7 at the time (Then K = 7)

Where we need to take in account the number of combinations and also the permutations for each combination.

Then the total number of different sets is:

C(15*7)*7!

First, the total number of combinations will be:

So we have 6,436 combinations, and each one of these combinations has 7! permutations.

permutations = 7! = 7*6*5*4*3*2*1 = 5,040

if we combine these we get:

Combinations*Permutations = 6,435*5,040 = 32,432,400