Code Password Generation Quiz: Code and Password Generation

The answer is True. P(26,26) = 26! divided by 0! = 26! divided by 1 = 26!. When all n symbols are used exactly once, the count equals n factorial — the total number of orderings of all 26 letters. Every distinct arrangement of all 26 letters is a unique 26-character code.

P(n,2) = n times (n-1) = 132. Solving: n squared minus n minus 132 = 0, giving (n-12)(n+11) = 0. The positive solution is n=12. Check: 12 times 11 = 132. Option A: 11 times 10 = 110, not 132. Option B: 13 times 12 = 156, not 132. Option C: 14 times 13 = 182, not 132. Only n=12 satisfies the equation.

P(36,4) = 1,413,720. 36P4 is standard notation confirming A. 36! divided by 32! = 36! divided by (36-4)! = P(36,4), confirming B. 36C4 times 4! = combinations times orderings = permutations, confirming C. 36 times 35 times 34 times 33 is the direct descending product, confirming D.

Choose the last digit from the 5 even digits (0,2,4,6,8): 5 ways. Fill the remaining 4 positions from the other 9 digits with no repetition: P(9,4) = 9 times 8 times 7 times 6 = 3024. Total = 5 times 3024 = 15,120. Option A gives 12,600, option B gives 13,860, option D gives 16,800, none of which correctly multiply 5 by P(9,4).

P(52,2) = 52 times 51 = 2652. The first position has 52 choices and the second has 51 since repetition is not allowed. Option A gives 2500 = 50 squared. Option B gives 2550 = 50 times 51. Option C gives 2600 = 52 times 50. Only 2652 correctly applies the no-repetition rule with n=52.

The answer is True. P(10,6) = 10 times 9 times 8 times 7 times 6 times 5 = 151,200. Verified: 10 times 9 = 90, times 8 = 720, times 7 = 5040, times 6 = 30,240, times 5 = 151,200. The statement correctly applies the permutation formula.

Since all 5 symbols are vowels, every 3-letter code without repetition automatically ends with a vowel. The total is simply P(5,3) = 5 times 4 times 3 = 60. Alternatively: choose the last position (5 ways) then arrange 2 from the remaining 4 (4 times 3 = 12), giving 5 times 12 = 60. Both methods confirm 60.

Option A: r=7 exceeds n=5, so no valid arrangements exist, confirming A. Option B: cannot select any symbol from an empty set, confirming B. Option D: need 7 distinct vowels but only 6 exist, so r exceeds n, confirming D. Option C: P(3,3) = 3! = 6, not 0 — using all 3 symbols once each produces valid codes.

Place A in one of 4 positions: 4 ways. Fill the remaining 3 positions from the other 11 symbols with no repetition: P(11,3) = 11 times 10 times 9 = 990. Total = 4 times 990 = 3960. Option A gives 3300, option B gives 3520, option C gives 3720, none of which correctly combine the position choices with P(11,3).

The first position is fixed (1 way), leaving 7 remaining symbols to be arranged in any order for positions 2 through 8. The number of arrangements is 7! = 5040. Option A gives 6! = 720. Option B gives 7 times 6 times 5 times ... stopped early. Option C gives 6 times 720 = 4320. Only 7! = 5040 correctly counts the arrangements of the remaining 7 symbols.

P(10,4) = 10 times 9 times 8 times 7 = 5040. The first position has 10 choices, second has 9, third has 8, fourth has 7. Option A gives 4096 = 4 to the power 6, unrelated. Option B gives 3024 = P(9,4), using only 9 digits. Option D gives 9072, which has no valid derivation.

P(10,3) = 720. 10P3 is standard notation for P(10,3), confirming A. 10 times 9 times 8 = 720, confirming B. 10! divided by 7! = 720, confirming C. 10C3 times 3! = 120 times 6 = 720, confirming D since multiplying combinations by r! recovers the permutation count.

P(36,5) = 36 times 35 times 34 times 33 times 32 = 45,239,040. Verified: 36 times 35 = 1,260; times 34 = 42,840; times 33 = 1,413,720; times 32 = 45,239,040. Options B, C, and D do not result from any valid application of the permutation formula with n=36 and r=5.

P(7,3) = 7 times 6 times 5 = 210. The first position has 7 choices, second has 6, third has 5. Option A gives 120 = P(5,3) or 5!. Option B gives 150, option C gives 180, neither of which result from the descending product starting at 7.

The answer is True. With no repetition, each position must use a distinct symbol. If more positions are needed than symbols available, it is impossible to fill all positions. The permutation formula breaks down since you would need to select more items than exist, giving a count of 0.

First position has 9 choices (digits 1 to 9, excluding 0). After choosing the first digit, 9 digits remain including 0, giving 9 choices for position 2. Then 8 for position 3 and 7 for position 4. Total = 9 times 9 times 8 times 7 = 4536. Option A uses 9 times 8 times 7 times 6, incorrectly excluding 0 from all positions. Option C ignores the restriction on the first digit.

P(26,4) = 26 times 25 times 24 times 23, confirming A and B. Since P(26,4) = 26! divided by (26-4)! = 26! divided by 22!, option C is also correct. Option D gives 26C4 = 14,950, which counts unordered combinations and is P(26,4) divided by 4!, not equal to the permutation count.

P(36,2) = 36 times 35 = 1260. The first position has 36 choices and the second has 35 since repetition is not allowed. Option A gives 1296 = 36 squared, which allows repetition. Options B and C have no valid derivation. Only 1260 correctly applies the no-repetition rule.

P(5,5) = 5! = 5 times 4 times 3 times 2 times 1 = 120. All 5 symbols must be used exactly once, so this counts all possible orderings of the 5 symbols. Option A gives 60 = P(5,3). Option B gives 80, option C gives 100, neither of which match 5!.

The answer is True. P(26,3) = 26 times 25 times 24 = 15,600. The first position has 26 choices, second has 25, third has 24, since no letter may repeat. The descending product formula for permutations gives exactly 15,600 ordered 3-letter codes.