Code Password Generation Quiz: Code And Password Generation

1) Using digits 0–9 with no repetition, how many 4-digit codes are possible?

4096

3024

5040

9072

This is permutations without repetition: P(10,4) = 10×9×8×7 = 5040.

Explanation

This is permutations without repetition: P(10,4) = 10×9×8×7 = 5040.

2) Using 26 uppercase letters with no repetition, the number of 3-letter codes is 26×25×24 = 15,600.

True

False

P(26,3) = 26×25×24 = 15,600, so the statement is true.

Explanation

P(26,3) = 26×25×24 = 15,600, so the statement is true.

Submit

4) From 36 symbols (A–Z and 0–9), how many 2-character codes without repetition?

1296

1350

1400

1260

P(36,2) = 36×35 = 1260.

Explanation

P(36,2) = 36×35 = 1260.

5) Select all expressions equal to the number of 4-letter codes from 26 uppercase letters without repetition.

26P4

26×25×24×23

26!/22!

26C4

26^4

Permutations: P(26,4) = 26×25×24×23 = 26!/22!. 26C4 is unordered; 26^4 allows repetition.

Explanation

Permutations: P(26,4) = 26×25×24×23 = 26!/22!. 26C4 is unordered; 26^4 allows repetition.

Submit

6) Digits 0–9, 4-character code, first position cannot be 0, no repetition. How many codes?

9×8×7×6 = 3024

9×9×8×7 = 4536

10×9×8×7 = 5040

9×10×9×8 = 6480

First digit: 1–9 (9 choices). After choosing first, remaining digits include 0, so positions 2–4 use 9,8,7 choices. Total 9×9×8×7 = 4536.

Explanation

First digit: 1–9 (9 choices). After choosing first, remaining digits include 0, so positions 2–4 use 9,8,7 choices. Total 9×9×8×7 = 4536.

7) If r>n, then the number of r-length codes from n symbols without repetition is 0.

True

False

You cannot place more distinct characters than available symbols, so the count is 0.

Explanation

You cannot place more distinct characters than available symbols, so the count is 0.

Submit

9) Using A–Z and 0–9 (36 symbols), how many 5-character codes without repetition?

45,239,040

41,140,800

36,288,000

34,650,240

P(36,5) = 36×35×34×33×32 = 45,239,040.

Explanation

P(36,5) = 36×35×34×33×32 = 45,239,040.

10) For 10 digits, select all expressions equal to the number of 3-character codes with no repetition.

10P3

10×9×8

10!/7!

10^3

10C3×3!

All listed except 10^3 (which allows repetition) equal P(10,3).

Explanation

All listed except 10^3 (which allows repetition) equal P(10,3).

Submit

11) Using 26 letters, the number of 26-character case-insensitive codes with no repetition is 26!.

True

False

Every permutation of the 26 letters is a unique 26-character code: P(26,26) = 26!.

Explanation

Every permutation of the 26 letters is a unique 26-character code: P(26,26) = 26!.

12) There are 8 distinct symbols. If the 8-character code must start with a specific symbol, the number of codes is ____.

Fix first position (1 way), then permute the other 7: 7! = 5040.

Explanation

Fix first position (1 way), then permute the other 7: 7! = 5040.

Submit

13) 12 distinct symbols, 4-character code must include symbol A exactly once (no repetition). How many such codes?

3300

3520

3720

3960

Place A in one of 4 positions (4 ways). Fill remaining 3 spots with P(11,3) = 11×10×9 = 990. Total 4×990 = 3960.

Explanation

Place A in one of 4 positions (4 ways). Fill remaining 3 spots with P(11,3) = 11×10×9 = 990. Total 4×990 = 3960.

14) Select all scenarios that result in 0 possible codes (no repetition).

N=5 symbols, r=7 characters

N=0 symbols, r=1 character

N=3 symbols, r=3 characters

Exactly 6 distinct vowels available, r=7 characters required

N=4 symbols, r=0 characters

Zero occurs when r>n or when constraints make selection impossible. A: 7>5 ⇒ 0. B: cannot choose 1 from 0 ⇒ 0. D: need 7 distinct vowels but only 6 exist ⇒ 0. C gives 3!; E gives 1.

Explanation

Zero occurs when r>n or when constraints make selection impossible. A: 7>5 ⇒ 0. B: cannot choose 1 from 0 ⇒ 0. D: need 7 distinct vowels but only 6 exist ⇒ 0. C gives 3!; E gives 1.

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15) Using only the 5 vowels A,E,I,O,U, how many 3-letter codes without repetition end with a vowel?

30

60

90

120

All letters are vowels. Choose the last (5 ways), then arrange 2 from remaining 4 for the first two spots: 4×3 = 12. Total 5×12 = 60 = P(5,3).

Explanation

All letters are vowels. Choose the last (5 ways), then arrange 2 from remaining 4 for the first two spots: 4×3 = 12. Total 5×12 = 60 = P(5,3).

16) Using digits 0–9 with no repetition, the number of 6-digit PINs is 10P6 = 151,200.

True

False

P(10,6) = 10×9×8×7×6×5 = 151,200.

Explanation

P(10,6) = 10×9×8×7×6×5 = 151,200.

Submit

18) Using digits 0–9 with no repetition, how many 5-character codes are even (last digit even)?

12,600

13,860

15,120

16,800

Choose last digit from {0,2,4,6,8} (5 ways). Fill remaining 4 positions with P(9,4) = 9×8×7×6 = 3024. Total 5×3024 = 15,120.

Explanation

Choose last digit from {0,2,4,6,8} (5 ways). Fill remaining 4 positions with P(9,4) = 9×8×7×6 = 3024. Total 5×3024 = 15,120.

19) From 36 symbols, select all expressions equal to the number of 4-character codes with no repetition.

36P4

36!/32!

36C4×4!

36×35×34×33

36^4

All except 36^4 (which allows repetition) match P(36,4).

Explanation

All except 36^4 (which allows repetition) match P(36,4).

Submit

20) If P(n, 2) = 132, what is n?

11

13

14

12

Solve n(n−1)=132 ⇒ n^2−n−132=0 ⇒ (n−12)(n+11)=0. Positive solution: n=12.

Explanation

Solve n(n−1)=132 ⇒ n^2−n−132=0 ⇒ (n−12)(n+11)=0. Positive solution: n=12.

Code Password Generation Quiz: Code and Password Generation