CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 01

Answer: (a) 20

Explanation:
1. A number between 500 and 600 is a 3-digit number with 5 in the hundred's place (1 way).
2. Ten's place: any of the remaining 5 digits; unit's place: any of the remaining 4 digits.
3. Required count = 1 × 5 × 4 = 20

Teacher's Note:
1. Always fix the most restricted place first; here the hundred's digit is forced to be 5.
2. Numbers like 500 or 600 cannot be formed anyway because 0 is not among the digits.

Answer: (a) 12

Explanation:
1. Onward journey: any of the 4 buses, so 4 ways.
2. Return journey must be by a different bus, so 3 ways.
3. Total = 4 × 3 = 12

Teacher's Note:
1. The word 'another' means the return bus must differ from the onward bus, so the choices drop from 4 to 3.
2. If the same bus were allowed for the return, the answer would be 4 × 4 = 16.

Answer: (b) 108

Explanation:
1. For an odd number the unit's digit must be 1, 3 or 5: 3 ways.
2. With repetition allowed, the hundred's and ten's places can each be filled in 6 ways.
3. Total = 6 × 6 × 3 = 108

Teacher's Note:
1. Start with the unit's place because the 'odd' condition restricts it.
2. When repetition is allowed, the choices for the other places do not decrease.

Answer: (b) 9000

Explanation:
1. The thousand's place can be any digit from 1 to 9: 9 ways.
2. Each of the other three places can be any digit from 0 to 9: 10 ways each.
3. Total = 9 × 10 × 10 × 10 = 9000

Teacher's Note:
1. A four-digit number cannot begin with 0, which is why the first place has only 9 choices.
2. Check: four-digit numbers run from 1000 to 9999, i.e. 9999 − 1000 + 1 = 9000 numbers.

Answer: (c) 9 × \( {}^{9}P_{3} \)

Explanation:
1. The first digit can be any of 1 to 9: 9 ways.
2. The remaining three places are filled by distinct digits from the 9 digits left (including 0): \( {}^{9}P_{3} \) ways.
3. Total = 9 × \( {}^{9}P_{3} \) = 9 × 504 = 4536

Teacher's Note:
1. Option \( {}^{10}P_{4} \) also counts arrangements starting with 0, which are not four-digit numbers.
2. After using one non-zero digit, 0 becomes available again, so 9 digits remain for the last three places.

Answer: (c) 4⁸

Explanation:
1. Each letter can be posted in any one of the 4 boxes: 4 ways for each letter.
2. The 8 letters are posted independently.
3. Total = 4 × 4 × … (8 times) = 4⁸

Teacher's Note:
1. Ask 'who chooses whom': each letter chooses a box, so the answer is (number of boxes)^{number of letters}.
2. 8⁴ would wrongly mean that each box chooses one of the letters.

Answer: (b) 48

Explanation:
1. For an even number the unit's digit must be 2 or 4: 2 ways.
2. The remaining 4 digits can be arranged in the other 4 places in 4! = 24 ways.
3. Total = 2 × 24 = 48

Teacher's Note:
1. Fill the restricted place (unit's digit) first and then arrange the rest freely.
2. As a check, 2 out of 5 digits are even, so 2/5 of all 120 arrangements are even: 48.

Answer: (c) 648

Explanation:
1. Hundred's place: digits 1–9 except 5, so 8 ways.
2. Ten's and unit's places: digits 0–9 except 5, so 9 ways each.
3. Total = 8 × 9 × 9 = 648

Teacher's Note:
1. Remember that 0 is excluded only from the first place.
2. The complement 900 − 648 = 252 gives the numbers having at least one digit 5.

Answer: (d) 365

Explanation:
1. Without 5: hundred's 8 ways (1–9 except 5), ten's 9 ways, unit's 5 ways (even): 8 × 9 × 5 = 360.
2. With 5: 5 cannot be in the unit's place (number is even) and must be followed by 7, so 5 cannot be in the ten's place either (unit's digit would be 7, odd).
3. So 5 is in the hundred's place with 7 in the ten's place: 57_ with an even unit's digit gives 5 numbers.
4. Total = 360 + 5 = 365

Teacher's Note:
1. Split into cases 'digit 5 absent' and 'digit 5 present' and handle each separately.
2. A condition like '7 must follow 5' quickly rules out positions where 5 cannot be followed by 7.

Answer: (a) 775

Explanation:
1. Count the three-digit numbers with all digits odd: each place can be 1, 3, 5, 7 or 9.
2. Such numbers = 5 × 5 × 5 = 125
3. Numbers with at least one even digit = 900 − 125 = 775

Teacher's Note:
1. 'At least one' problems are usually easiest by subtracting the 'none' case from the total.
2. Zero is an even digit, so numbers like 105 are included in the count.

Answer: (d) 720

Explanation:
1. The even digits are 2, 4, 6. The first and last places can be filled with even digits in \( {}^{3}P_{2} \) = 6 ways.
2. The middle 4 places are filled from the remaining 5 digits in \( {}^{5}P_{4} \) = 120 ways.
3. Total = 6 × 120 = 720

Teacher's Note:
1. 'Terminal digits' means the first and the last digits of the number.
2. Only 6 of the 7 digits are used, so the middle places use a permutation \( {}^{5}P_{4} \), not 5!.

Answer: (a) 12 × 81

Explanation:
1. The first strip can be coloured in 4 ways.
2. Each of the next 5 strips can take any colour except that of the strip just before it: 3 ways each.
3. Total = 4 × 3⁵ = 972 = 12 × 81

Teacher's Note:
1. For 'no two adjacent alike' problems, only the immediately preceding strip restricts the choice.
2. Write the result in the form of the options: 4 × 243 = 972 = 12 × 81.

Answer: (b) 45000

Explanation:
1. Fill the first four digits freely: 9 × 10 × 10 × 10 = 9000 ways.
2. For any choice, exactly 5 of the 10 possible last digits make the total sum even.
3. Total = 9000 × 5 = 45000

Teacher's Note:
1. Fixing the last digit by parity is a neat trick: whatever the partial sum, half the digits fix the parity.
2. This also shows that exactly half of all 90000 five-digit numbers have an even digit sum.

Answer: (c) 21

Explanation:
1. The two non-zero digits have sum 8: {1,7}, {2,6}, {3,5}, {4,4}.
2. The first digit is non-zero; the other non-zero digit can be in any of the remaining 3 places (zeros fill the rest): 3 positions.
3. Distinct pairs {1,7}, {2,6}, {3,5}: 2 orders × 3 positions = 6 numbers each, i.e. 18; pair {4,4}: 3 numbers.
4. Total = 18 + 3 = 21

Teacher's Note:
1. Treat the pair {4, 4} separately, because swapping identical digits does not give a new number.
2. Placing the leading non-zero digit first automatically avoids numbers beginning with 0.

Answer: (a) 16800

Explanation:
1. INDEPENDENCE has 12 letters: vowels E, E, E, E, I (5) and consonants N, N, N, D, D, P, C (7).
2. Treat the 5 vowels as one unit: 8 units with N thrice and D twice: \( \frac{8!}{3!\,2!} = 3360 \) ways.
3. The vowels inside the unit can be arranged in \( \frac{5!}{4!} = 5 \) ways.
4. Total = 3360 × 5 = 16800

Teacher's Note:
1. Remember to divide by the factorials of repeated letters both outside and inside the block.
2. The block method works for any 'always together' condition.

Answer: (b) 432,000

Explanation:
1. Girls together: treat them as one unit with 6 boys: 7! × 5! = 604800 arrangements.
2. Of these, the boys are also all together when the arrangement is (girls block)(boys block) in either order: 2 × 6! × 5! = 172800.
3. Required = 604800 − 172800 = 432000

Teacher's Note:
1. When a condition says 'but not', count the main condition first and subtract the unwanted cases.
2. Both blocks together can be arranged only in 2 orders, which gives the subtracted term.

Answer: (c) 309

Explanation:
1. Alphabetical order of letters: E, H, M, O, R, T.
2. Words beginning with E or H: 2 × 5! = 240. Words beginning ME, MH: 2 × 4! = 48. Words beginning MOE, MOH, MOR: 3 × 3! = 18.
3. Words beginning MOTE: 2! = 2. Next comes MOTHER itself (MOTHER before MOTHRE).
4. Rank = 240 + 48 + 18 + 2 + 1 = 309

Teacher's Note:
1. For dictionary rank, count all words that start with a smaller letter at each position, then add 1.
2. Always write the letters in alphabetical order first to avoid missing any case.

Answer: (a) 614

Explanation:
1. Alphabetical order: A, D, M, N, O, R.
2. Words starting with A, D, M, N, O: 5 × 5! = 600.
3. R A _ : RAD, RAM before RAN: 2 × 3! = 12. RAN D M O… : RANDMO comes before RANDOM: 1.
4. Words before RANDOM = 600 + 12 + 1 = 613, so rank = 614

Teacher's Note:
1. At each position count only the unused letters smaller than the actual letter.
2. Do not forget the final +1: the rank is (number of words before it) + 1.

Answer: (c) 9

Explanation:
1. Lines joining 6 vertices = \( {}^{6}C_{2} \) = 15.
2. Of these, 6 are the sides.
3. Diagonals = 15 − 6 = 9

Teacher's Note:
1. Number of diagonals of an n-gon = \( \frac{n(n-3)}{2} \); for n = 6 this gives 9.
2. Each vertex sends n − 3 diagonals, and every diagonal is counted from both ends.

Answer: (a) 13

Explanation:
1. Diagonals: \( \frac{n(n-3)}{2} = 65 \), so n(n − 3) = 130.
2. n² − 3n − 130 = 0 ⇒ (n − 13)(n + 10) = 0
3. n = 13 (n cannot be negative)

Teacher's Note:
1. Set up the diagonal formula and solve the quadratic; reject the negative root.
2. Quick check: 13 × 10 / 2 = 65.

Answer: (c) 54

Explanation:
1. Each exterior angle = 180° − 150° = 30°, so the number of sides n = 360°/30° = 12.
2. Diagonals = \( \frac{12(12-3)}{2} = 54 \)

Teacher's Note:
1. Working with exterior angles is quicker because they always add up to 360°.
2. Then use \( \frac{n(n-3)}{2} \) for the diagonals.

Answer: (b) 2 · \( {}^{20}C_{2} \)

Explanation:
1. Each pair of students exchanges 2 cards (one each way).
2. Number of pairs = \( {}^{20}C_{2} \) = 190.
3. Cards = 2 × \( {}^{20}C_{2} \) = 380 (which also equals \( {}^{20}P_{2} \))

Teacher's Note:
1. Card sending is an ordered relation (A to B differs from B to A), so it is a permutation \( {}^{20}P_{2} \) = 2 · \( {}^{20}C_{2} \).
2. Handshakes, in contrast, are unordered and equal \( {}^{20}C_{2} \).

Answer: (c) 56

Explanation:
1. Each visit corresponds to a different group (selection) of 3 children out of 8.
2. Number of visits = \( {}^{8}C_{3} \) = 56

Teacher's Note:
1. The order in which the children are taken does not matter, so use combinations, not permutations.
2. 336 = \( {}^{8}P_{3} \) would count ordered groups and is a common wrong answer.

Answer: (c) 9

Explanation:
1. Two black balls: \( {}^{4}C_{2} \) = 6 ways.
2. Two white balls: \( {}^{3}C_{2} \) = 3 ways.
3. Total = 6 + 3 = 9

Teacher's Note:
1. The cases 'both black' and 'both white' cannot happen together, so add them.
2. Use multiplication only when choices are made one after another within the same case.

Answer: (b) 720

Explanation:
1. Choose 3 bowlers out of 5: \( {}^{5}C_{3} \) = 10; 1 wicket-keeper out of 2: \( {}^{2}C_{1} \) = 2.
2. The remaining 7 players come from the other 16 − 5 − 2 = 9 players: \( {}^{9}C_{7} \) = 36.
3. Total = 10 × 2 × 36 = 720

Teacher's Note:
1. 'Exactly three bowlers and one wicket-keeper' means the other 7 must be chosen only from the remaining 9 players.
2. Split the group into separate categories and multiply the selections from each.

Answer: (d) 8

Explanation:
1. No couple in the committee means exactly one person from each of the 3 couples.
2. From each couple choose husband or wife: 2 ways.
3. Total = 2 × 2 × 2 = 8

Teacher's Note:
1. With 3 members and 3 couples, avoiding couples forces one member from every couple.
2. Check: total committees \( {}^{6}C_{3} \) = 20, and those containing a couple are 3 × 4 = 12; 20 − 12 = 8.

Answer: (c) 135

Explanation:
1. Total people = 20, so all possible handshakes = \( {}^{20}C_{2} \) = 190.
2. Indian wives do not shake hands with any of the 10 males: 5 × 10 = 50 handshakes removed.
3. American wives do not shake hands with their own husbands: 5 more removed.
4. Handshakes = 190 − 50 − 5 = 135

Teacher's Note:
1. Start from all possible handshakes and subtract the forbidden pairs, taking care not to subtract any pair twice.
2. Indian husband–wife pairs are already included in the 50 removed, so only the American couples are removed separately.

Answer: (d) 80

Explanation:
1. No complete pair means the 4 shoes come from 4 different pairs: \( {}^{5}C_{4} \) = 5 ways to choose the pairs.
2. From each chosen pair take the left or the right shoe: 2⁴ = 16 ways.
3. Total = 5 × 16 = 80

Teacher's Note:
1. First choose which pairs are used, then which shoe of each pair.
2. Choosing shoe by shoe (10 × 8 × 6 × 4) counts ordered selections; divide by 4! to get 80 again.

Answer: (a) 624

Explanation:
1. Possible splits of 5 questions with at least one from each part: (3, 1, 1) and (2, 2, 1) in some order.
2. (3, 1, 1): 3 choices of the part with 3 questions × \( {}^{4}C_{3} \) × 4 × 4 = 3 × 64 = 192.
3. (2, 2, 1): 3 choices of the part with 1 question × \( {}^{4}C_{2} \) × \( {}^{4}C_{2} \) × 4 = 3 × 144 = 432.
4. Total = 192 + 432 = 624

Teacher's Note:
1. Choosing one from each part first and then 2 more from the rest overcounts; splitting into cases avoids this.
2. Remember to multiply by the number of ways to decide which part gets which share.

Answer: (b) 10; (c) 11

Explanation:
1. The number of different groups of size r is \( {}^{21}C_{r} \).
2. For odd n = 21, \( {}^{21}C_{r} \) is greatest at r = 10 and r = 11 (equal values, since \( {}^{21}C_{10} \) = \( {}^{21}C_{11} \)).
3. So the group size is 10 or 11.

Teacher's Note:
1. \( {}^{n}C_{r} \) is maximum at r = n/2 for even n and at both middle values r = (n − 1)/2, (n + 1)/2 for odd n.
2. This follows from \( {}^{n}C_{r} \) = \( {}^{n}C_{n-r} \), which makes the values symmetric.

Answer: (c) 1560

Explanation:
1. d is the largest digit. Choose 7 distinct digits; once chosen, their positions are decided by choosing which 3 of the other 6 go to the left (in increasing order): \( {}^{6}C_{3} \) = 20 ways.
2. Without the zero condition: \( {}^{10}C_{7} \) × 20 = 120 × 20 = 2400.
3. Invalid ones have a = 0: then 0 is on the left, and choose 6 more digits from 1–9 (\( {}^{9}C_{6} \) = 84) and 2 of the other 5 for the left (\( {}^{5}C_{2} \) = 10): 840.
4. Required = 2400 − 840 = 1560

Teacher's Note:
1. Once a set of digits is chosen, an increasing or decreasing order fixes their arrangement in exactly one way.
2. Zero can never be the leading digit, so subtract the cases where 0 lands in position a.

Answer: (b) 160

Explanation:
1. The grid has 7 vertical and 5 horizontal lines (6 × 4 unit squares). Rectangles = \( {}^{7}C_{2} \) × \( {}^{5}C_{2} \) = 21 × 10 = 210.
2. Squares: 1 × 1: 24, 2 × 2: 15, 3 × 3: 8, 4 × 4: 3; total 50.
3. Rectangles that are not squares = 210 − 50 = 160

Teacher's Note:
1. A rectangle is fixed by choosing 2 vertical and 2 horizontal lines.
2. For a k × k square in an m × n grid, the count is (m − k + 1)(n − k + 1).

Answer: (a) \( {}^{n}C_{2} \)

Explanation:
1. No three points on a circle are collinear.
2. Each pair of points gives a distinct line, so the number of lines = \( {}^{n}C_{2} \).

Teacher's Note:
1. A straight line meets a circle in at most two points, so no three of the points are collinear.
2. Lines are unordered pairs of points, hence a combination, not a permutation.

Answer: (c) 270

Explanation:
1. Line–line: \( {}^{9}C_{2} \) × 1 = 36.
2. Circle–circle: \( {}^{9}C_{2} \) × 2 = 72.
3. Line–circle: 9 × 9 × 2 = 162.
4. Total = 36 + 72 + 162 = 270

Teacher's Note:
1. Two lines meet in at most 1 point, two circles in at most 2, and a line and a circle in at most 2.
2. For the maximum, assume no three curves pass through the same point.

Answer: (d) 711

Explanation:
1. Each side contains 7 of the points (5 marked points plus its two vertices).
2. Total selections of 3 points = \( {}^{18}C_{3} \) = 816.
3. Collinear triples on each side = \( {}^{7}C_{3} \) = 35, for 3 sides: 105.
4. Triangles = 816 − 105 = 711

Teacher's Note:
1. The vertices A, B, C lie on two sides each, so count the points of every side including its end vertices.
2. Subtract the selections of 3 points from the same side because they do not form a triangle.

Answer: (a) 100

Explanation:
1. All selections of 3 points = \( {}^{10}C_{3} \) = 120.
2. Selections from the 6 collinear points = \( {}^{6}C_{3} \) = 20 (no triangle).
3. Triangles = 120 − 20 = 100

Teacher's Note:
1. Use the result: triangles = \( {}^{n}C_{3} \) − \( {}^{m}C_{3} \) when m of the n points are collinear.
2. Assume no other three of the points are collinear.

Answer: (d) 185

Explanation:
1. Lines from 20 points = \( {}^{20}C_{2} \) = 190.
2. The 4 collinear points give \( {}^{4}C_{2} \) = 6 pairs but only 1 line.
3. Lines = 190 − 6 + 1 = 185

Teacher's Note:
1. Remove the overcounted lines through collinear points and add back the single line they actually form.
2. Formula: \( {}^{n}C_{2} \) − \( {}^{m}C_{2} \) + 1.

Answer: (b) 8

Explanation:
1. Number of pentagons = \( {}^{n}C_{5} \), number of triangles = \( {}^{n}C_{3} \).
2. \( {}^{n}C_{5} \) = \( {}^{n}C_{3} \) ⇒ 5 + 3 = n (since 5 ≠ 3)
3. n = 8

Teacher's Note:
1. If \( {}^{n}C_{x} \) = \( {}^{n}C_{y} \), then x = y or x + y = n.
2. Points on a circle are never three collinear, so every selection gives a valid polygon.

Answer: (c) n²(4n − 3)

Explanation:
1. Total points = 3n; selections of 3 = \( {}^{3n}C_{3} \).
2. Collinear selections on each line: \( {}^{n}C_{3} \), three lines: 3 · \( {}^{n}C_{3} \).
3. \( {}^{3n}C_3 - 3\,{}^{n}C_3 = \frac{3n(3n-1)(3n-2) - 3n(n-1)(n-2)}{6} = n^2(4n-3) \)

Teacher's Note:
1. Points on the same line cannot form a triangle; subtract them line by line.
2. Check with n = 1: three points, one on each line, form exactly 1 triangle and n²(4n − 3) = 1.

Answer: (a) \( \frac{m+n-2}{m+n} \)

Explanation:
1. (i) Triangles: 2 points on AB and 1 on AC, or 1 on AB and 2 on AC: \( {}^{m}C_2\,n + m\,{}^{n}C_2 = \frac{mn(m+n-2)}{2} \).
2. (ii) With A included, add triangles with vertex A and one point on each line: mn.
3. Case (ii) = \( \frac{mn(m+n-2)}{2} + mn = \frac{mn(m+n)}{2} \)
4. Ratio = (m + n − 2) : (m + n)

Teacher's Note:
1. Every triangle must use points from both lines, otherwise its vertices would be collinear.
2. Adding A creates extra triangles only of the type A with one point from each line.

Answer: (a) 106

Explanation:
1. Line–line: \( {}^{7}C_{2} \) − \( {}^{3}C_{2} \) = 21 − 3 = 18.
2. Circle–circle: 2(\( {}^{5}C_{2} \) − 1) = 2 × 9 = 18 (the concentric pair does not meet).
3. Line–circle: 7 × 5 × 2 = 70.
4. Total = 18 + 18 + 70 = 106

Teacher's Note:
1. Parallel lines do not intersect and concentric circles do not intersect, so remove those pairs.
2. Each remaining pair of circles and each line–circle pair can give 2 points.

Answer: (a) 2880

Explanation:
1. Seat the 5 boys: 5! = 120 ways.
2. A girl between two boys must sit in one of the 4 inner gaps (not at the ends): \( {}^{4}P_{3} \) = 24 ways.
3. Total = 120 × 24 = 2880

Teacher's Note:
1. 'Between two boys' excludes the two end gaps, unlike 'no two girls together'.
2. Count gaps carefully: 5 boys create 4 inner gaps and 2 end gaps.

Answer: (c) 96

Explanation:
1. Total 3-letter words from 6 letters = \( {}^{6}P_{3} \) = 120.
2. Words with no vowel (using only b, c, d, f) = \( {}^{4}P_{3} \) = 24.
3. Required = 120 − 24 = 96

Teacher's Note:
1. The vowels here are a and e; complement counting avoids messy cases.
2. 'At least one' is almost always handled as Total − None.