NCERT Solutions for Class 11 Maths Chapter 6: Permutations and Combinations (NCERT 2026–27)

These Class 11 Maths Chapter 6 solutions cover Permutations and Combinations from the NCERT textbook (Reprint 2026–27). Every question of Exercise 6.1, 6.2, 6.3, 6.4 and the Miscellaneous Exercise is solved step by step — using the fundamental principle of counting, the formulas nPr and nCr, and factorial notation — with every final answer verified against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 6 Exercises: 6.1, 6.2, 6.3, 6.4 & Miscellaneous Topic: Permutations and Combinations Session: 2026–27
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Chapter 6 Overview

Chapter 6, Permutations and Combinations, teaches you to count arrangements and selections without listing them one by one. It begins with the fundamental principle of counting (the multiplication principle), then introduces permutations — arrangements where order matters — together with factorial notation and the formula nPr = n!/(n − r)!. It also handles permutations when some objects are alike (e.g. repeated letters), and finishes with combinations — selections where order does not matter — using nCr = n!/[r!(n − r)!]. The Class 11 Maths Chapter 6 solutions below work through every exercise question step by step.

Key Concepts & Definitions

Fundamental principle of counting: if one event can occur in m ways and, following it, another can occur in n ways, the two together occur in m × n ways (extends to any number of events).

Factorial: n! = 1 × 2 × 3 × … × n is the product of the first n natural numbers, with the convention 0! = 1.

Permutation: an arrangement, in a definite order, of a number of objects taken some or all at a time — order matters.

Combination: a selection of objects where the order does not matter; choosing X and Y is the same as choosing Y and X.

Distinct vs. repeated: when all objects are different the count uses nPr; when some are alike (p of one kind, q of another…) the arrangements reduce to n!/(p! q! …).

Relation: permutations and combinations are linked by nPr = nCr × r!, since each combination of r objects can itself be arranged in r! ways.

Important Formulas (Chapter 6)

Multiplication principle: total ways = m × n × p × … for events in succession.

Factorial: n! = n × (n − 1)! ,  0! = 1.

Permutations (no repetition): nPr = n!/(n − r)! ,  0 ≤ r ≤ n;   nPn = n!.

Permutations (repetition allowed): nr.

Permutations with alike objects: n!/(p1! p2! … pk!).

Combinations: nCr = n!/[r!(n − r)!] ,  0 ≤ r ≤ n.

Useful identities: nCr = nCn−r;   nCr + nCr−1 = n+1Cr;   nPr = nCr × r!.

Exercise 6.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?

SOLUTION A 3-digit number has three places: hundreds, tens, units. (i) Repetition allowed: each of the 3 places can be filled by any of the 5 digits, so total = 5 × 5 × 5 = 125. (ii) No repetition: hundreds in 5 ways, tens in 4 ways, units in 3 ways → 5 × 4 × 3 = 60.

2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

SOLUTION For the number to be even, the units place must be 2, 4 or 6 → 3 choices. Repetition is allowed, so hundreds and tens can each be any of the 6 digits. Total = 6 (hundreds) × 6 (tens) × 3 (units) = 108.

3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

SOLUTION There are 4 places and 10 available letters (A–J) with no repetition. Total = 10 × 9 × 8 × 7 = 10P4 = 5040.

4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

SOLUTION The first two digits are fixed as 6 and 7, so the digits 6 and 7 are used up. The remaining 3 places are filled from the other 8 digits with no repetition. Total = 8 × 7 × 6 = 8P3 = 336.

5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

SOLUTION Each toss has 2 outcomes (Head or Tail), and the three tosses are independent. Total = 2 × 2 × 2 = 23 = 8.

6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

SOLUTION Two ordered places are to be filled from 5 different flags (order matters since one is below the other). Upper place in 5 ways, lower place in 4 ways → 5 × 4 = 5P2 = 20.

Exercise 6.2 Solutions

1. Evaluate (i) 8 ! (ii) 4 ! – 3 !

SOLUTION (i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320. (ii) 4! = 24 and 3! = 6, so 4! − 3! = 24 − 6 = 18.

2. Is 3 ! + 4 ! = 7 ! ?

SOLUTION 3! + 4! = 6 + 24 = 30. 7! = 5040. Since 30 ≠ 5040, the statement is No (false).

3. Compute 8! / (6! × 2!)

SOLUTION 8! / (6! × 2!) = (8 × 7 × 6!)/(6! × 2) = (8 × 7)/2 = 56/2 = 28.

4. If 1/6! + 1/7! = x/8! , find x.

SOLUTION Write 7! = 7 × 6! and 8! = 8 × 7 × 6!. So 1/6! + 1/(7 × 6!) = x/(8 × 7 × 6!). Multiply throughout by 8 × 7 × 6!: (8 × 7) + 8 = x. x = 56 + 8 = 64.

5. Evaluate n! / (n − r)! , when (i) n = 6, r = 2 (ii) n = 9, r = 5.

SOLUTION (i) 6!/(6 − 2)! = 6!/4! = 6 × 5 = 30. (ii) 9!/(9 − 5)! = 9!/4! = 9 × 8 × 7 × 6 × 5 = 15120.

Exercise 6.3 Solutions

1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

SOLUTION Arrange 9 distinct digits in 3 places with no repetition (order matters). Total = 9P3 = 9 × 8 × 7 = 504.

2. How many 4-digit numbers are there with no digit repeated?

SOLUTION Digits available are 0–9 (ten digits). The thousands place cannot be 0, so it has 9 choices. After fixing the first digit, the remaining 3 places are filled from the other 9 digits: 9 × 8 × 7. Total = 9 × 9 × 8 × 7 = 4536.

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

SOLUTION For an even number the units digit must be 2, 4 or 6 → 3 choices. After fixing the units digit, the remaining two places are filled from the remaining 5 digits: 5 × 4 = 20. Total = 3 × (5 × 4) = 3 × 20 = 60.

4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

SOLUTION Total 4-digit numbers: arrange 5 distinct digits in 4 places = 5P4 = 5 × 4 × 3 × 2 = 120. Even numbers: units digit must be 2 or 4 → 2 choices. The remaining 3 places are filled from the remaining 4 digits: 4 × 3 × 2 = 24. Even numbers = 2 × 24 = 48.

5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

SOLUTION The two posts are distinct, so order matters → this is a permutation. Chairman in 8 ways, vice chairman in 7 ways → 8P2 = 8 × 7 = 56.

6. Find n if n−1P3 : nP4 = 1 : 9.

SOLUTION n−1P3 / nP4 = [(n − 1)!/(n − 4)!] ÷ [n!/(n − 4)!] = (n − 1)!/n! = 1/n. So 1/n = 1/9 ⇒ n = 9.

7. Find r if (i) 5Pr = 2 6Pr−1   (ii) 5Pr = 6Pr−1.

SOLUTION (i) 5!/(5 − r)! = 2 × 6!/(6 − r + 1)! = 2 × 6!/(7 − r)!. (7 − r)! = (7 − r)(6 − r)(5 − r)!, so dividing: 1 = 2 × 6 / [(7 − r)(6 − r)] ⇒ (7 − r)(6 − r) = 12. r2 − 13r + 42 = 12 ⇒ r2 − 13r + 30 = 0 ⇒ (r − 3)(r − 10) = 0. Since r ≤ 5, take r = 3. (ii) 5!/(5 − r)! = 6!/(7 − r)! ⇒ (7 − r)(6 − r) = 6!/5! = 6. r2 − 13r + 42 = 6 ⇒ r2 − 13r + 36 = 0 ⇒ (r − 4)(r − 9) = 0. Since r ≤ 5, take r = 4.

8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

SOLUTION EQUATION has 8 distinct letters (E, Q, U, A, T, I, O, N). Arranging all 8 = 8! = 40320.

9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel?

SOLUTION MONDAY has 6 distinct letters; vowels are O and A. (i) 6P4 = 6 × 5 × 4 × 3 = 360. (ii) All 6 letters = 6! = 720. (iii) Fix a vowel at the first place (2 choices: O or A), then arrange the remaining 5 letters in 5! = 120 ways → 2 × 120 = 240.

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

SOLUTION MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2). Total distinct arrangements = 11!/(4! 4! 2!) = 39916800/(24 × 24 × 2) = 34650. Arrangements with all four I’s together: treat the four I’s as one block, leaving 8 items M, S, S, S, S, P, P and (IIII): 8!/(4! 2!) = 40320/(24 × 2) = 840. Required (I’s not together) = 34650 − 840 = 33810.

11. In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?

SOLUTION PERMUTATIONS has 12 letters in which T appears twice and all the rest are different; vowels are E, U, A, I, O (5 vowels). (i) Fix P at the start and S at the end. The remaining 10 letters (with T repeated twice) are arranged in 10!/2! = 3628800/2 = 1814400 ways. (ii) Treat the 5 vowels (E, U, A, I, O) as one block. This block with the 7 consonants (P, R, M, T, T, N, S) gives 8 items, with T repeated twice: 8!/2! = 20160 arrangements. The 5 vowels inside the block (all distinct) arrange in 5! = 120 ways. Total = 20160 × 120 = 2419200. (iii) “4 letters between P and S” means P and S occupy positions (1,6), (2,7), (3,8), (4,9), (5,10), (6,11) or (7,12) — 7 position-pairs — and P, S can be swapped, giving 7 × 2 = 14 ways to place P and S. The remaining 10 letters (T repeated twice) fill the other 10 places in 10!/2! = 1814400 ways. Total = 14 × 1814400 = 25401600.

Exercise 6.4 Solutions

1. If nC8 = nC2, find nC2.

SOLUTION If nCa = nCb then a = b or a + b = n. Here 8 ≠ 2, so n = 8 + 2 = 10. 10C2 = (10 × 9)/(2 × 1) = 45.

2. Determine n if (i) 2nC3 : nC3 = 12 : 1 (ii) 2nC3 : nC3 = 11 : 1

SOLUTION 2nC3 / nC3 = [2n(2n − 1)(2n − 2)/6] ÷ [n(n − 1)(n − 2)/6] = [2n(2n − 1) · 2(n − 1)] / [n(n − 1)(n − 2)] = 4(2n − 1)/(n − 2). (i) 4(2n − 1)/(n − 2) = 12 ⇒ 2n − 1 = 3(n − 2) = 3n − 6 ⇒ n = 5. (ii) 4(2n − 1)/(n − 2) = 11 ⇒ 8n − 4 = 11n − 22 ⇒ 3n = 18 ⇒ n = 6.

3. How many chords can be drawn through 21 points on a circle?

SOLUTION A chord is fixed by choosing 2 points; order does not matter. Number of chords = 21C2 = (21 × 20)/2 = 210.

4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

SOLUTION Choose 3 boys from 5 and 3 girls from 4 (selection, so combinations). 5C3 × 4C3 = 10 × 4 = 40.

5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

SOLUTION Choose 3 red from 6, 3 white from 5, 3 blue from 5. 6C3 × 5C3 × 5C3 = 20 × 10 × 10 = 2000.

6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

SOLUTION Choose exactly 1 ace from the 4 aces and the remaining 4 cards from the 48 non-ace cards. 4C1 × 48C4 = 4 × 194580 = 778320.

7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

SOLUTION There are 5 bowlers and 12 non-bowlers (17 − 5). The team needs exactly 4 bowlers and 7 non-bowlers. 5C4 × 12C7 = 5 × 792 = 3960.

8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

SOLUTION Choose 2 black from 5 and 3 red from 6. 5C2 × 6C3 = 10 × 20 = 200.

9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

SOLUTION Two courses are compulsory, so they are already chosen. The student must pick the remaining 3 courses from the other 7. 7C3 = (7 × 6 × 5)/(3 × 2 × 1) = 35.

Miscellaneous Exercise on Chapter 6 — Solutions

1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?

SOLUTION DAUGHTER has 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R). Select 2 vowels from 3: 3C2 = 3. Select 3 consonants from 5: 5C3 = 10. Selections = 3 × 10 = 30. Each selection has 5 letters that can be arranged in 5! = 120 ways. Total words = 30 × 120 = 3600.

2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

SOLUTION EQUATION has 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). Treat all vowels as one block (V) and all consonants as one block (C). These 2 blocks can be arranged in 2! = 2 ways. Within the vowel block: 5! = 120 arrangements. Within the consonant block: 3! = 6 arrangements. Total = 2! × 5! × 3! = 2 × 120 × 6 = 1440.

3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?

SOLUTION (i) Exactly 3 girls → 3 girls and 4 boys: 4C3 × 9C4 = 4 × 126 = 504. (ii) At least 3 girls → (3 girls, 4 boys) or (4 girls, 3 boys): 4C3×9C4 + 4C4×9C3 = 504 + (1 × 84) = 588. (iii) At most 3 girls → 0, 1, 2 or 3 girls: 4C0×9C7 + 4C1×9C6 + 4C2×9C5 + 4C3×9C4 = (1×36) + (4×84) + (6×126) + (4×126) = 36 + 336 + 756 + 504 = 1632.

4. If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E ?

SOLUTION EXAMINATION has 11 letters: A(2), I(2), N(2), E, X, M, T, O. Words before the first word starting with E are exactly those starting with A. Fix A at the first place. The remaining 10 letters are A(1), I(2), N(2), E, X, M, T, O — with I and N each repeated twice. Number of such arrangements = 10!/(2! 2!) = 3628800/4 = 907200.

5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?

SOLUTION For divisibility by 10 the units digit must be 0, so fix 0 in the units place. The remaining 5 places are filled by the other 5 digits (1, 3, 5, 7, 9) with no repetition: 5! = 120. Total = 120.

6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?

SOLUTION Select 2 vowels from 5: 5C2 = 10. Select 2 consonants from 21: 21C2 = 210. Selections = 10 × 210 = 2100. Each selection has 4 different letters that arrange in 4! = 24 ways. Total words = 2100 × 24 = 50400.

7. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ?

SOLUTION Part I has 5 questions, Part II has 7; attempt 8 with at least 3 from each. Possible (I, II) splits: (3,5), (4,4), (5,3). (3,5): 5C3 × 7C5 = 10 × 21 = 210. (4,4): 5C4 × 7C4 = 5 × 35 = 175. (5,3): 5C5 × 7C3 = 1 × 35 = 35. Total = 210 + 175 + 35 = 420.

8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

SOLUTION Choose exactly 1 king from the 4 kings and the other 4 cards from the remaining 48 non-king cards. Number of ways = 4C1 × 48C4 = 4 × 194580 = 778320.

9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?

SOLUTION There are 9 seats. The even places are positions 2, 4, 6, 8 — exactly 4 places — for the 4 women: 4! = 24 ways. The 5 men take the 5 odd places (1, 3, 5, 7, 9): 5! = 120 ways. Total = 5! × 4! = 120 × 24 = 2880.

10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?

SOLUTION Case 1 — the 3 join: 3 are fixed, choose the remaining 7 from the other 22 students: 22C7. Case 2 — the 3 do not join: choose all 10 from the other 22 students: 22C10. Total ways = 22C7 + 22C10 (= 170544 + 646646 = 817190).

11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ?

SOLUTION ASSASSINATION has 13 letters: A(3), S(4), I(2), N(2), T(1), O(1). Treat the four S’s as one single block. The items to arrange are then: (SSSS), A, A, A, I, I, N, N, T, O — 10 items, with A repeated 3 times, I twice, N twice. Number of arrangements = 10!/(3! 2! 2!) = 3628800/(6 × 2 × 2) = 3628800/24 = 151200.

Common Mistakes to Avoid

Watch out for these

  • Confusing permutations (order matters — arrangements, posts, words, numbers) with combinations (order does not matter — teams, committees, chords, selections).
  • Forgetting that the leading digit cannot be 0 in counting how many numbers can be formed.
  • Dividing by the factorial of repeats: for words with alike letters use n!/(p! q! …), and after fixing letters, re-count remaining repeats correctly.
  • In “together” problems, remembering to multiply by the internal arrangements of the block (e.g. 5! for the vowel block).
  • For “at least / at most” problems, listing all the valid cases and adding them — do not stop at one case.
  • Using nCa = nCb ⇒ a = b only; the other possibility a + b = n is often the intended one.

Practice MCQs & Assertion–Reason

1. The value of 5! is:

(a) 25    (b) 60    (c) 120    (d) 720

2. The number of ways of arranging 4 different books on a shelf is:

(a) 4    (b) 16    (c) 24    (d) 256

3. 10P2 equals:

(a) 20    (b) 45    (c) 90    (d) 100

4. The number of chords that can be drawn through 21 points on a circle is:

(a) 21    (b) 42    (c) 210    (d) 441

5. If nC8 = nC2, then n is:

(a) 6    (b) 8    (c) 10    (d) 16

6. The number of 3-digit numbers from digits 1–9 with no digit repeated is:

(a) 504    (b) 729    (c) 720    (d) 84

7. The number of ways to select 3 boys and 3 girls from 5 boys and 4 girls is:

(a) 20    (b) 40    (c) 60    (d) 120

8. The number of words formed using all letters of EQUATION (each once) is:

(a) 5040    (b) 40320    (c) 720    (d) 362880

9. The relation between permutations and combinations is:

(a) nPr = nCr    (b) nPr = nCr × r!    (c) nCr = nPr × r!    (d) nPr = nCr + r!

10. The number of distinct arrangements of the letters of MISSISSIPPI is:

(a) 11!    (b) 34650    (c) 33810    (d) 840

Answer key: 1-(c), 2-(c), 3-(c), 4-(c), 5-(c), 6-(a), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The number of ways to choose a chairman and a vice chairman from 8 persons is 56.

Reason: Selecting two persons for two distinct posts is a permutation, 8P2 = 56.

A-R 2. Assertion: nCr = nCn−r.

Reason: Choosing r objects out of n is the same as rejecting the remaining (n − r) objects.

A-R 3. Assertion: The number of chords through 21 points on a circle is 21P2.

Reason: A chord joins two points and the order of the two points matters.

A-R 4. Assertion: 0! = 1.

Reason: There is exactly one way of arranging zero objects.

A-R 5. Assertion: The number of arrangements of all letters of EQUATION so that vowels and consonants occur together is 1440.

Reason: The vowel block (5!) and consonant block (3!) can themselves be ordered in 2! ways, giving 2! × 5! × 3! = 1440.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Multiplication principle: events in succession multiply — m × n × …
  • Factorial: n! = 1 × 2 × … × n, with 0! = 1.
  • Permutation (order matters): nPr = n!/(n − r)!; all at a time nPn = n!; with repetition allowed nr.
  • Alike objects: arrangements = n!/(p1! p2! … pk!).
  • Combination (order does not matter): nCr = n!/[r!(n − r)!].
  • Key identities: nCr = nCn−r; nCr + nCr−1 = n+1Cr; nPr = nCr × r!.
  • “Together”: tie objects into a block, arrange the blocks, then multiply by internal arrangements.

How to score full marks in this chapter

First decide whether order matters — that tells you permutation vs. combination. For numbers, watch the leading-zero condition; for words, divide by the factorials of repeated letters. Solve “at least / at most” problems by splitting into clear cases and adding, and solve “together” problems with the block method, never forgetting the internal arrangements. Write the formula you are using and show each substitution so every step earns its mark.

Frequently Asked Questions

What is Class 11 Maths Chapter 6 about?

Chapter 6, Permutations and Combinations, covers the fundamental principle of counting, factorial notation, permutations (arrangements where order matters, using nPr), permutations of objects that are not all distinct, and combinations (selections where order does not matter, using nCr).

How many exercises are there in Class 11 Maths Chapter 6?

There are four exercises — Exercise 6.1, 6.2, 6.3 and 6.4 — plus a Miscellaneous Exercise on Chapter 6. Every question of all five sets is solved step by step on this page.

What is the difference between a permutation and a combination?

A permutation is an arrangement in which the order matters (e.g. forming words or filling distinct posts), while a combination is a selection in which the order does not matter (e.g. forming a team or committee). They are linked by nPr = nCr × r!.

Are these Class 11 Maths Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every final answer verified against the book’s answer key.

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