How many three-letter (unordered) sets are possible that use the letters q, u, a, k, e, s at most once each? (No Response) 20 sets

20sets

Step-by-step explanation:

Since we are to select 3 unordered letters from the word q, u, a, k, e, s, we will apply the combination rule.

For example if r objects are to be selected from n pool of objects, this can be done in nCr number of ways.

nCr = n!/(n-r)!r!

Since the total number of letter in the word q, u, a, k, e, s is 6letters and we are to form 3 letters from it unordered, this can be done in 6C3 number of ways.

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 6×5×4×3×2×1/3×2×1×3×2×1

6C3 = 6×5×4/3×2

6C3 = 120/6

6C3 = 20

Hence 20sets of selection are possible

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.