How many ways are there to arrange m A’s, n B’s and 1 C in a circle on a piece of paper? Two letter arrangements are equivalent if and only if one can be rotated to obtain the other without flipping the paper over. Clearly and concisely explain how you got your answer

The number of ways are;

(m+n)!/m!n!

Step-by-step explanation:

In this question , we are asked to calculate the number of ways in which we can make an arrangement of letters in a circle on a piece of paper.

We proceed as follows;

Number of ways in which we can arrange k letters in a circle on a piece of paper without flipping = (k-1)!

In this case there are m A's , n B's and 1 C. So, k = m + n + 1.

But m A's are identical to each other. Similarly, n B's are identical to each other.

Hence, number of ways in which we can arrange m A's , n B's and 1 C in a circle on a piece of paper

= {(k-1)!}/{m!n!}

= {(m+n+1)-1)!}/{m!n!}

= {(m+n)!}/{m!n!}

no this is a nonlinear negative

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: