Permutations with Repetition Quiz � Distinct Multiset Arrangements (HSS.CP.B.9)

1) Find the number of distinct permutations of the letters in "LEVEL".

30

36

24

40

“LEVEL” has 5 letters with L repeated 2 times and E repeated 2 times: 5!/(2!·2!) = 120/4 = 30.

Explanation

“LEVEL” has 5 letters with L repeated 2 times and E repeated 2 times: 5!/(2!·2!) = 120/4 = 30.

2) Find the number of distinct permutations of the letters in “BALLOON”.

1260

1440

1080

1500

“BALLOON” has 7 letters with L repeated 2 times and O repeated 2 times: 7!/(2!·2!) = 5040/4 = 1260.

Explanation

“BALLOON” has 7 letters with L repeated 2 times and O repeated 2 times: 7!/(2!·2!) = 5040/4 = 1260.

3) How many 5-digit numbers can be formed using the digits 1,1,1,2,3?

20

24

18

30

Five symbols with 1 repeated thrice: 5!/3! = 120/6 = 20.

Explanation

Five symbols with 1 repeated thrice: 5!/3! = 120/6 = 20.

4) Find the number of distinct arrangements of the letters in “SUCCESS”.

420

504

360

480

“SUCCESS” has 7 letters with S repeated 3 times and C repeated 2 times: 7!/(3!·2!) = 5040/12 = 420.

Explanation

“SUCCESS” has 7 letters with S repeated 3 times and C repeated 2 times: 7!/(3!·2!) = 5040/12 = 420.

5) Find the number of distinct arrangements of the letters in “TATTOO”.

60

200

150

210

“TATTOO” has 6 letters with T repeated 3 times and O repeated 2 times: 6!/(3!·2!) = 720/12 = 60.

Explanation

“TATTOO” has 6 letters with T repeated 3 times and O repeated 2 times: 6!/(3!·2!) = 720/12 = 60.

6) Find the number of arrangements of the letters in “MISSISSIPPI”.

34,650

36,000

32,400

35,000

“MISSISSIPPI” has 11 letters: I=4, S=4, P=2 → 11!/(4!·4!·2!) = 39,916,800 / 1,152 = 34,650.

Explanation

“MISSISSIPPI” has 11 letters: I=4, S=4, P=2 → 11!/(4!·4!·2!) = 39,916,800 / 1,152 = 34,650.

7) Find the number of arrangements of the letters in “MOON”.

12

16

18

20

“MOON” has 4 letters with O repeated twice: 4!/2! = 24/2 = 12.

Explanation

“MOON” has 4 letters with O repeated twice: 4!/2! = 24/2 = 12.

8) Find the number of 6-letter arrangements possible from “BANANA”.

60

72

48

80

“BANANA” has 6 letters with A=3 and N=2: 6!/(3!·2!) = 720/12 = 60.

Explanation

“BANANA” has 6 letters with A=3 and N=2: 6!/(3!·2!) = 720/12 = 60.

9) Find the number of arrangements of the digits 2,2,3,3,3.

10

12

15

20

Five digits with 2 repeated twice and 3 repeated thrice: 5!/(2!·3!) = 120/12 = 10.

Explanation

Five digits with 2 repeated twice and 3 repeated thrice: 5!/(2!·3!) = 120/12 = 10.

10) Find the number of arrangements of the letters in “BOOKKEEPER”.

151,200

160,000

140,000

155,000

“BOOKKEEPER” has 10 letters with O=2, K=2, E=3: 10!/(2!·2!·3!) = 3,628,800 / 24 = 151,200.

Explanation

“BOOKKEEPER” has 10 letters with O=2, K=2, E=3: 10!/(2!·2!·3!) = 3,628,800 / 24 = 151,200.

Permutations with Repetition — Basics