Permutation problem that i just do not feel like solving your cousin, who is planning her wedding, is working on the seating chart for the reception. she is trying to decide which 6 people should be seated at the table closest to the head table. she has narrowed her decision down to a list of 10 friends. if the order doesn't matter, in how many ways can she choose 6 friends from the list of 10 to sit at the table closest to the head table?

210. she has 210 ways to choose 6 friends from the list of 10 to sit at the table closest to the head table no matter the order.

This is a problem of combinations and can be solved using the equation , where n! and k! is the factorial of a number. The factorial is defined in principle as the product of all positive integers from 1 (ie, natural numbers) to n.

She has a list of 10 friends and we want to know in how many ways she can choose 6 friends.

Using the combinations equation, with n = 10 and k = 6:

the answer is 8

step-by-step explanation:

3.14 x 8x8

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°