CBSE Class 11 Mathematics Permutations and Combinations JEE Previous Year Questions Set 01

Answer: (b) 196

Explanation:
1. Case 4 from the first five and 6 from the other 8: \( {}^{5}C_{4} \) × \( {}^{8}C_{6} \) = 5 × 28 = 140.
2. Case all 5 from the first five and 5 from the other 8: \( {}^{5}C_{5} \) × \( {}^{8}C_{5} \) = 56.
3. Total = 140 + 56 = 196

Teacher's Note:
1. Break 'at least 4' into the exact cases 4 and 5.
2. Choosing 4 first and then 6 from the remaining 9 would double count, so avoid that shortcut.

Answer: (b) \( {}^{n+2}C_{r+1} \)

Explanation:
1. Write the sum as \( ({}^{n}C_{r+1} + {}^{n}C_{r}) + ({}^{n}C_{r} + {}^{n}C_{r-1}) \).
2. By \( {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r} \): the sum = \( {}^{n+1}C_{r+1} + {}^{n+1}C_{r} \).
3. Applying the identity again gives \( {}^{n+2}C_{r+1} \).

Teacher's Note:
1. The identity \( {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r} \) is used twice; split 2·\( {}^{n}C_{r} \) into two equal parts.
2. Check with n = 2, r = 1: 1 + 1 + 4 = 6 = \( {}^{4}C_{2} \).

Answer: (a) {1, 2, 3}

Explanation:
1. For \( {}^{n}P_{r} \) we need r ≥ 0 and n ≥ r: x − 3 ≥ 0 and 7 − x ≥ x − 3, so 3 ≤ x ≤ 5 (x an integer).
2. f(3) = \( {}^{4}P_{0} \) = 1, f(4) = \( {}^{3}P_{1} \) = 3, f(5) = \( {}^{2}P_{2} \) = 2.
3. Range = {1, 2, 3}

Teacher's Note:
1. First find the domain from the conditions on n and r, then evaluate at each allowed value.
2. Remember \( {}^{n}P_{0} = 1 \) for every n ≥ 0.

Answer: (c) 360

Explanation:
1. GARDEN has 6 distinct letters: 6! = 720 arrangements.
2. In exactly half of them the vowels A and E appear in the order A before E.
3. Required = 720/2 = 360

Teacher's Note:
1. Fixing the order of k particular letters divides the total by k!.
2. Alternatively choose 2 places for the vowels (\( {}^{6}C_{2} \) = 15) and arrange the 4 consonants (24): 360.

Answer: (d) 601

Explanation:
1. Alphabetical order: A, C, H, I, N, S. Words beginning with A, C, H, I, N: 5 × 5! = 600.
2. The first word beginning with S is SACHIN itself (remaining letters A, C, H, I, N in order).
3. Serial number = 600 + 1 = 601

Teacher's Note:
1. Here every letter after S is already in alphabetical order, so SACHIN is the first S-word.
2. Counting the blocks of 5! words for each smaller first letter is the key step.

Answer: (b) 385

Explanation:
1. He may vote for 1, 2, 3 or 4 candidates.
2. Ways = \( {}^{10}C_{1} \) + \( {}^{10}C_{2} \) + \( {}^{10}C_{3} \) + \( {}^{10}C_{4} \) = 10 + 45 + 120 + 210 = 385

Teacher's Note:
1. Different numbers of votes are separate cases, so add the combinations.
2. Order of voting is irrelevant, hence combinations.

Answer: (a) 7 · \( {}^{6}C_{4} \) · \( {}^{8}C_{4} \)

Explanation:
1. Arrange M, I, I, I, I, P, P (7 letters): \( \frac{7!}{4!\,2!} = 105 = 7 \cdot {}^{6}C_4 \).
2. These create 8 gaps; place the 4 identical S's in 4 different gaps: \( {}^{8}C_{4} \) ways.
3. Total = 7 · \( {}^{6}C_{4} \) · \( {}^{8}C_{4} \) = 105 × 70 = 7350

Teacher's Note:
1. Gap method again: arrange the other letters, then put the S's in separate gaps.
2. Rewrite 105 as 7 × 15 = 7 · \( {}^{6}C_{4} \) to match the options.

Answer: (c) at least 1000

Explanation:
1. Select 4 novels: \( {}^{6}C_{4} \) = 15; select 1 dictionary: \( {}^{3}C_{1} \) = 3.
2. Dictionary fixed in the middle; the 4 novels are arranged in the other 4 places: 4! = 24.
3. Total = 15 × 3 × 24 = 1080, which is at least 1000.

Teacher's Note:
1. First select, then arrange; the dictionary's position is fixed so only novels are permuted.
2. This question appeared in both 2009 and 2018 with the same answer.

Answer: (b) N ≤ 100

Explanation:
1. N = \( {}^{10}C_{3} \) − \( {}^{6}C_{3} \) = 120 − 20 = 100.
2. So N ≤ 100.

Teacher's Note:
1. Subtract the selections of 3 collinear points from all selections.
2. N = 100 exactly, so the inequality must include equality.

Answer: (b) 5

Explanation:
1. \( T_n = {}^{n}C_3 \), so \( T_{n+1} - T_n = {}^{n+1}C_3 - {}^{n}C_3 = {}^{n}C_2 \).
2. \( {}^{n}C_2 = 10 \) ⇒ n(n − 1) = 20 ⇒ n = 5

Teacher's Note:
1. Use \( {}^{n+1}C_3 = {}^{n}C_3 + {}^{n}C_2 \) to simplify the difference.
2. Vertices of a polygon are never three collinear, so every triple forms a triangle.

Answer: (b) 108

Explanation:
1. There are 4! = 24 such numbers.
2. Each digit occurs in the unit's place 3! = 6 times.
3. Sum of unit digits = 6 × (3 + 4 + 5 + 6) = 6 × 18 = 108

Teacher's Note:
1. By symmetry every digit appears equally often in each place.
2. The same idea gives the sum of all the numbers: 108 × 1111.

Answer: (d) 36 (7!)

Explanation:
1. Sum of 0 to 9 is 45; leaving out two digits must keep the sum divisible by 9, so the omitted pair sums to 9.
2. Pairs: (0, 9), (1, 8), (2, 7), (3, 6), (4, 5).
3. Omitting (0, 9): digits 1–8, 8! = 8(7!) numbers. Omitting any other pair: 0 is present, so 8! − 7! = 7(7!) each; 4 pairs give 28(7!).
4. Total = 8(7!) + 28(7!) = 36(7!)

Teacher's Note:
1. A number is divisible by 9 exactly when the sum of its digits is divisible by 9.
2. Whenever 0 is among the digits, remove the arrangements that start with 0.

Answer: (b) 120

Explanation:
1. Odd digits 1, 1, 3 must go in even places (2, 4, 6, 8): choose 3 of the 4 even places and arrange 1, 1, 3: \( {}^{4}C_{3} \) × \( \frac{3!}{2!} \) = 4 × 3 = 12.
2. Remaining 5 places get 2, 2, 2, 4, 4: \( \frac{5!}{3!\,2!} = 10 \).
3. Total = 12 × 10 = 120

Teacher's Note:
1. Place the restricted (odd) digits first, then fill the remaining places.
2. Use the alike-objects formula for both stages because of repeated digits.

Answer: (b) [10, 12)

Explanation:
1. Let there be m men. Games among men = 2 · \( {}^{m}C_{2} \) = m(m − 1).
2. Games between men and women = 2 × 2 × m = 4m.
3. m(m − 1) − 4m = 66 ⇒ m² − 5m − 66 = 0 ⇒ (m − 11)(m + 6) = 0 ⇒ m = 11
4. 11 lies in the interval [10, 12).

Teacher's Note:
1. Every pair plays two games, so each pair count is doubled.
2. Read interval brackets carefully: (11, 13] does not contain 11.

Answer: (d) 192

Explanation:
1. 4-digit numbers greater than 6000 must start with 6, 7 or 8: 3 × \( {}^{4}P_{3} \) = 3 × 24 = 72.
2. All 5-digit numbers exceed 6000: 5! = 120.
3. Total = 72 + 120 = 192

Teacher's Note:
1. Do not forget the 5-digit numbers; they are automatically greater than 6000.
2. No 0 is available, so every arrangement is a valid number.

Answer: (c) 15 × 14!

Explanation:
1. Each of the 15 teams pairs one man with one woman, and all 30 persons are used.
2. Pair the men in a fixed order with the women: the first man has 15 choices, the next 14, and so on: 15! ways.
3. 15! = 15 × 14!

Teacher's Note:
1. Forming all the teams together is like matching two sets of equal size, which gives n! ways.
2. Write the answer in the form given in the options: 15! = 15 × 14!.

Answer: (a) 12

Explanation:
1. \( \frac{n(n-3)}{2} = 54 \) ⇒ n² − 3n − 108 = 0
2. (n − 12)(n + 9) = 0 ⇒ n = 12

Teacher's Note:
1. Diagonals of an n-gon = \( {}^{n}C_2 - n = \frac{n(n-3)}{2} \).
2. Check: 12 × 9 / 2 = 54.

Answer: (c) 58^th

Explanation:
1. Letters: A, L, L, M, S. Words starting with A: \( \frac{4!}{2!} = 12 \); with L: 4! = 24; with M: 12. Total 48.
2. Starting with SA: \( \frac{3!}{2!} = 3 \); with SL: 3! = 6. Total so far 57.
3. The next word is SMALL, so its position is 58.

Teacher's Note:
1. When the first letter is L, only one L remains, so the other four letters are all different.
2. After SL, the next block SM begins with SMALL itself because A, L, L are already in order.

Answer: (d) 59

Explanation:
1. MEDITERRANEAN: E×3, R×2, A×2, N×2, M, D, I, T. After using one R and one E, the available letters are E×2, R×1, A×2, N×2, M, D, I, T (8 kinds).
2. Second and third letters different: 8 × 7 = 56.
3. Second and third letters the same: possible only for E, A, N: 3.
4. Total = 56 + 3 = 59

Teacher's Note:
1. After fixing letters, recount what is still available before filling the middle places.
2. Treat 'both letters same' separately because it needs a letter available at least twice.

Answer: (a) n² + 3n − 108 = 0

Explanation:
1. \( \frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{720 (n-2)(n-3)} = 11 \) ⇒ (n + 2)(n + 1)n(n − 1) = 7920.
2. 7920 = 11 × 10 × 9 × 8, so n + 2 = 11, i.e. n = 9.
3. n = 9 satisfies 81 + 27 − 108 = 0, i.e. n² + 3n − 108 = 0.

Teacher's Note:
1. Cancel the common factors \( (n-2)(n-3) \) before solving.
2. Write the product as consecutive integers to spot n quickly.

Answer: (a) 485

Explanation:
1. If X invites k ladies and (3 − k) men, Y must invite (3 − k) ladies and k men.
2. k = 0: 1 · 1 · 1 · 1 = 1; k = 1: \( {}^{4}C_{1} \)·\( {}^{3}C_{2} \)·\( {}^{3}C_{2} \)·\( {}^{4}C_{1} \) = 144.
3. k = 2: \( {}^{4}C_{2} \)·\( {}^{3}C_{1} \)·\( {}^{3}C_{1} \)·\( {}^{4}C_{2} \) = 324; k = 3: \( {}^{4}C_{3} \)·1·1·\( {}^{4}C_{3} \) = 16.
4. Total = 1 + 144 + 324 + 16 = 485

Teacher's Note:
1. Make the cases depend on one variable (ladies invited by X) so that both conditions hold automatically.
2. Within each case, multiply the four independent selections.

Answer: (c) 46^th

Explanation:
1. Letters: E, E, N, Q, U. Words starting with E: 4! = 24; with N: \( \frac{4!}{2!} = 12 \). Total 36.
2. Starting with QE: 3! = 6; with QN: \( \frac{3!}{2!} = 3 \). Total 45.
3. The first word starting with QU is QUEEN, so its position is 46.

Teacher's Note:
1. When one E is used up, the remaining letters may become all distinct; adjust the formula each time.
2. Add 1 at the end because QUEEN itself is the next word.

Answer: (b) 7

Explanation:
1. Each of the n places can be filled with 2, 5 or 7: 3ⁿ numbers.
2. Need 3ⁿ ≥ 900: 3⁶ = 729 < 900, 3⁷ = 2187 ≥ 900.
3. Smallest n = 7

Teacher's Note:
1. With repetition allowed, each place is independent, giving 3ⁿ.
2. Compare successive powers of 3 with 900 to find the least n.

Answer: (d) 270

Explanation:
1. BARRACK: A×2, R×2, B, C, K (5 kinds).
2. All different: \( {}^{5}P_{4} \) = 120.
3. One pair alike + 2 different: 2 choices of pair × \( {}^{4}C_{2} \) × \( \frac{4!}{2!} \) = 2 × 6 × 12 = 144.
4. Two pairs (AARR): \( \frac{4!}{2!\,2!} = 6 \). Total = 120 + 144 + 6 = 270

Teacher's Note:
1. Split by the pattern of repeated letters: all different, one pair, two pairs.
2. After choosing a pair, pick the other two letters from the remaining kinds only.

Answer: (b) 30

Explanation:
1. Sum of 0 + 1 + 2 + 3 + 4 = 10; omit one digit so the remaining sum is a multiple of 3: omit 1 or 4.
2. Digits {0, 2, 3, 4}: first digit 2, 3 or 4: 3 × 3! = 18.
3. Digits {0, 1, 2, 3}: first digit 2 or 3: 2 × 3! = 12.
4. Total = 18 + 12 = 30

Teacher's Note:
1. Divisibility by 3 depends only on the digit sum, so first decide which digits are used.
2. The first digit must be 2, 3 or 4 to lie between 2000 and 5000.

Answer: (a) 69760

Explanation:
1. All 5-letter words with repetition allowed = 10⁵ = 100000.
2. Words with all letters different = \( {}^{10}P_{5} \) = 30240.
3. Words with at least one letter repeated = 100000 − 30240 = 69760

Teacher's Note:
1. 'At least one repeated' is the complement of 'all different'.
2. 30240 (option b) is the count of words with no repetition, a common trap.

Answer: (c) \( {}^{52}C_{4} \)

Explanation:
1. The sum is \( {}^{47}C_4 + {}^{47}C_3 + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 \).
2. \( {}^{47}C_4 + {}^{47}C_3 = {}^{48}C_4 \), then \( {}^{48}C_4 + {}^{48}C_3 = {}^{49}C_4 \), and so on.
3. Finally \( {}^{51}C_4 + {}^{51}C_3 = {}^{52}C_4 \).

Teacher's Note:
1. Write the terms in increasing order of the upper index so that \( {}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r \) can be applied repeatedly.
2. j = 5 gives the term \( {}^{47}C_{3} \), which starts the chain.

Answer: (d) None of these

Explanation:
1. Women (distinct people) take 2 of the chairs 1–4 in order: \( {}^{4}P_{2} \) = 12 ways.
2. The men then take 3 of the remaining 6 chairs: \( {}^{6}P_{3} \) = 120 ways.
3. Total = 12 × 120 = 1440, which matches none of (a), (b), (c).

Teacher's Note:
1. People are distinct, so assigning chairs is a permutation, not a combination.
2. After the women sit, 6 chairs (not 4) remain for the men.

Answer: (a) 216

Explanation:
1. Sum of 0 to 5 is 15; leaving out one digit, the remaining sum is divisible by 3 only if the digit left out is 0 or 3.
2. Digits 1, 2, 3, 4, 5: 5! = 120 numbers.
3. Digits 0, 1, 2, 4, 5: 5! − 4! = 96 numbers.
4. Total = 120 + 96 = 216

Teacher's Note:
1. Divisibility by 3 depends on the digit sum, so choose the digit set first.
2. Remove the arrangements beginning with 0 when 0 is in the set.

Answer: (c) 60

Explanation:
1. Odd digits 3, 3, 5, 5 go to the 4 even positions: \( \frac{4!}{2!\,2!} = 6 \) ways.
2. Even digits 2, 2, 8, 8, 8 go to the 5 odd positions: \( \frac{5!}{2!\,3!} = 10 \) ways.
3. Total = 6 × 10 = 60

Teacher's Note:
1. There are exactly 4 even positions and 4 odd digits, so the odd digits fill them completely.
2. Use the alike-objects formula separately for each group.

Answer: (a) 40

Explanation:
1. Total arrangements = \( \frac{6!}{3!\,2!} = 60 \).
2. N's together (treat NN as one unit): \( \frac{5!}{3!} = 20 \).
3. N's not adjacent = 60 − 20 = 40

Teacher's Note:
1. Not adjacent = Total − Adjacent.
2. Gap method check: arrange B, A, A, A (4 ways), then choose 2 of 5 gaps for N's (10): 40.

Answer: (c) 96

Explanation:
1. Letters: C, C, H, I, N, O. Words beginning with C and then C, H, I or N: 4 × 4! = 96.
2. The first word beginning with CO is COCHIN itself (C, H, I, N already in order).
3. Words before COCHIN = 96

Teacher's Note:
1. Only words starting with C can come before COCHIN, since C is the smallest letter.
2. Within the C-block, count by the second letter.

Answer: (c) 77

Explanation:
1. Seven digits each at least 1 give sum at least 7; the extra 3 must come from 2's (extra 1) and 3's (extra 2).
2. One 3 and one 2 with five 1's: \( \frac{7!}{5!} = 42 \).
3. Three 2's with four 1's: \( \frac{7!}{3!\,4!} = 35 \).
4. Total = 42 + 35 = 77

Teacher's Note:
1. Thinking of the 'extra over 1' simplifies the possible digit patterns.
2. Each pattern is then arranged using the alike-objects formula.

Answer: (d) 126

Explanation:
1. S has 5 odd and 4 even numbers, so every 5-element subset contains at least 1 odd number.
2. Hence the sum counts all 5-element subsets: \( {}^{9}C_{5} \) = 126.

Teacher's Note:
1. Recognise that the cases k = 1 to 5 together cover every 5-element subset.
2. No 5-element subset can have 0 odd numbers because there are only 4 even numbers.

Answer: (a) 380

Explanation:
1. No boy: \( {}^{6}C_{4} \) = 15 teams; one boy: \( {}^{6}C_{3} \) × \( {}^{4}C_{1} \) = 80 teams. Total 95 teams.
2. A captain is chosen from the 4 members: 4 ways.
3. Total = 95 × 4 = 380

Teacher's Note:
1. Remember the captain: the 95 teams must be multiplied by 4.
2. 'At most one boy' means 0 boys or 1 boy.

Answer: (a) (A)→(p), (B)→(s), (C)→(q), (D)→(q)

Explanation:
1. ENDEANOEL: E×3, N×2, D, A, O, L. (A) ENDEA as one unit with N, O, E, L: 5 distinct units, 5! → (p).
2. (B) Middle 7 letters N, N, D, E, A, O, L: \( \frac{7!}{2!} = 2520 = 21 \times 5! \) → (s).
3. (C) Last five must be E, E, E, A, O: \( \frac{5!}{3!} = 20 \); first four D, L, N, N: \( \frac{4!}{2!} = 12 \); 240 = 2 × 5! → (q).
4. (D) A, E, E, E, O fill the 5 odd positions: 20 ways; N, N, D, L the even ones: 12 ways; 240 = 2 × 5! → (q).

Teacher's Note:
1. List the letter counts of the word first; every part depends on the repeated E's and N's.
2. Express each answer as a multiple of 5! to match Column II.

Answer: (a) (A)→(S), (B)→(U), (C)→(T), (D)→(Q)

Explanation:
1. α₁ = \( {}^{6}C_{3} \) × \( {}^{5}C_{2} \) = 20 × 10 = 200.
2. α₂ = Σ (k boys, k girls), k = 1 to 5: 6·5 + 15·10 + 20·10 + 15·5 + 6·1 = 30 + 150 + 200 + 75 + 6 = 461.
3. α₃ = \( {}^{11}C_{5} \) − (0 girls) − (1 girl) = 462 − \( {}^{6}C_{5} \) − \( {}^{6}C_{4} \)·\( {}^{5}C_{1} \) = 462 − 6 − 75 = 381.
4. α₄: 4 members with ≥ 2 girls = \( {}^{5}C_{2} \)·\( {}^{6}C_{2} \) + \( {}^{5}C_{3} \)·6 + \( {}^{5}C_{4} \) = 150 + 60 + 5 = 215; with both M₁, G₁: other 2 from 5 boys and 4 girls with at least one girl = \( {}^{9}C_{2} \) − \( {}^{5}C_{2} \) = 26; α₄ = 215 − 26 = 189.

Teacher's Note:
1. For 'at least' conditions, subtract the small unwanted cases from the total.
2. For 'not together', subtract the committees containing both persons.

Answer: 35

Explanation:
1. Arrange the six identical '+' signs: 1 way; they create 7 gaps.
2. Place the four identical '−' signs in 4 different gaps: \( {}^{7}C_{4} \) = 35 ways.

Teacher's Note:
1. Gap method with identical objects uses combinations for choosing gaps.
2. Since all '+' signs are alike and all '−' signs are alike, no factorial factor appears.

Answer: 625

Explanation:
1. The last two digits must form a number divisible by 4: 12, 24, 32, 44, 52 (5 ways).
2. The first three digits can each be any of 5 digits: 5³ = 125.
3. Total = 125 × 5 = 625

Teacher's Note:
1. Divisibility by 4 is decided by the last two digits only.
2. With repetition allowed, the other places are completely free.

Answer: 9

Explanation:
1. \( \frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} = \frac{84}{36} = \frac{7}{3} \) ⇒ 3n − 10r + 3 = 0.
2. \( \frac{{}^{n}C_{r+1}}{{}^{n}C_r} = \frac{n-r}{r+1} = \frac{126}{84} = \frac{3}{2} \) ⇒ 2n − 5r − 3 = 0.
3. Solving: r = 3 and n = 9.

Teacher's Note:
1. The ratio \( \frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} \) turns such problems into linear equations.
2. Check: \( {}^{9}C_{2} \) = 36, \( {}^{9}C_{3} \) = 84, \( {}^{9}C_{4} \) = 126.

Answer: 64

Explanation:
1. Total ways to draw 3 of the 9 balls = \( {}^{9}C_{3} \) = 84.
2. Draws with no black ball (from the 6 others) = \( {}^{6}C_{3} \) = 20.
3. Required = 84 − 20 = 64

Teacher's Note:
1. Balls are treated as distinct here, so use combinations of 9 objects.
2. 'At least one black' = Total − No black.

Answer: 6062

Explanation:
1. Women w = 5 to 9 with 12 − w men: \( {}^{9}C_{5} \)·\( {}^{8}C_{7} \) + \( {}^{9}C_{6} \)·\( {}^{8}C_{6} \) + \( {}^{9}C_{7} \)·\( {}^{8}C_{5} \) + \( {}^{9}C_{8} \)·\( {}^{8}C_{4} \) + \( {}^{9}C_{9} \)·\( {}^{8}C_{3} \).
2. = 1008 + 2352 + 2016 + 630 + 56
3. = 6062 (of these, women are in majority in 2016 + 630 + 56 = 2702 and men in majority in 1008)

Teacher's Note:
1. With 8 men only, at least 4 women are needed anyway, so the case list starts at w = 5.
2. Organise the cases in a table of women and men to avoid missing any.

Answer: 5

Explanation:
1. n: girls as one block with 5 boys: 6! × 5! ways.
2. m: arrange boys (5!), creating 6 gaps; choose and order 4 girls for a block (\( {}^{5}P_{4} \) = 120) and put the block and the remaining girl in different gaps (6 × 5 = 30): m = 5! × 120 × 30.
3. \( \frac{m}{n} = \frac{5! \times 3600}{6! \times 5!} = \frac{3600}{720} = 5 \)

Teacher's Note:
1. 'Exactly four together' means the fifth girl must be separated from the block, which the gap method ensures.
2. Keep common factors like 5! unsimplified until the final ratio.

Answer: 5

Explanation:
1. x = 10! (all ten letters used once).
2. y: choose the repeated letter (10 ways), choose 8 other letters from the remaining 9 (\( {}^{9}C_{8} \) = 9), arrange the 10 letters: \( \frac{10!}{2!} \).
3. \( y = 10 \times 9 \times \frac{10!}{2} = 45 \times 10! \), so \( \frac{y}{9x} = \frac{45}{9} = 5 \)

Teacher's Note:
1. With one letter used twice, only 9 distinct letters appear, so one letter of the 10 is left out.
2. Divide by 2! because the repeated letter's two copies are identical.