Advanced Permutations Quiz � Constraints, Blocks, and Mixed Orders (HSS.CP.B.9)

1) How many 5-letter arrangements can be made from the letters in the word "ALGEBRA" if no letter repeats?

12

2520

720

360

(7 distinct letters total, so 7P5 = 7×6×5×4×3 = 2520.)

Explanation

(7 distinct letters total, so 7P5 = 7×6×5×4×3 = 2520.)

2) How many ways can 4 students be seated in a row of 8 chairs?

1680

420

336

5040

(8P4 = 8×7×6×5 = 1680.)

Explanation

(8P4 = 8×7×6×5 = 1680.)

3) How many different lineups of 5 runners are possible if there are 9 total runners?

15,120

6,048

3,024

18,000

(9P5 = 9×8×7×6×5 = 15,120.)

Explanation

(9P5 = 9×8×7×6×5 = 15,120.)

4) A security code consists of 3 different letters followed by 2 different digits. How many codes are possible?

25,000

6,50,000

15,60,000

23,40,000

(26P3 × 10P2 = 26×25×24 × 10×9 = 1,560,000.)

Explanation

(26P3 × 10P2 = 26×25×24 × 10×9 = 1,560,000.)

5) How many ways can 4 different paintings be arranged on a wall if two specific paintings must stay together as a unit?

12

48

24

36

(Treat the 2 as one unit → 3 objects total → 3! = 6 ways. But the pair can switch order internally → ×2 = 12.)

Explanation

(Treat the 2 as one unit → 3 objects total → 3! = 6 ways. But the pair can switch order internally → ×2 = 12.)

6) How many 5-letter codes can be formed from the alphabet if repetition is allowed?

120

1,18,81,376

8,000

26,000

(26⁵ = 11,881,376.)

Explanation

(26⁵ = 11,881,376.)

7) A locker combination uses 3 letters followed by 3 digits. Letters cannot repeat, but digits can. How many combinations are possible?

15,81,84,000

1,40,400

15,60,000

2,00,00,000

(26P3 × 10³ = 26×25×24 × 1000 = 158,184,000.)

Explanation

(26P3 × 10³ = 26×25×24 × 1000 = 158,184,000.)

8) From a group of 10 people, how many different 6-person lineups can be formed?

90,000

720

3,00,000

1,51,200

(10P6 = 10×9×8×7×6×5 = 151,200.)

Explanation

(10P6 = 10×9×8×7×6×5 = 151,200.)

9) How many different ways can the letters in the word “ALGEBRA” be arranged?

210

2520

5040

720

(The word ALGEBRA has 7 letters with “A” repeated twice → 7!/2! = 5040/2 = 2520.)

Explanation

(The word ALGEBRA has 7 letters with “A” repeated twice → 7!/2! = 5040/2 = 2520.)

10) How many different ways can 6 students stand in a single line if the tallest student must be in the middle?

120

720

60

360

(Fix the tallest in position 3. That leaves 5 students to arrange in the remaining 5 spots: 5! = 120.)

Explanation

(Fix the tallest in position 3. That leaves 5 students to arrange in the remaining 5 spots: 5! = 120.)

Applying Permutations in Context (Advanced)

What do you like, dislike, and how can we improve?