JEE Mathematics Permutation and Combination MCQs Set 12

Answer: (c) 15

Explanation:
1. One flag: 3 ways; two flags: \( 3\times2=6 \) ways; three flags: \( 3\times2\times1=6 \) ways.
2. Total \( 3+6+6=15 \).

Teacher's Note:
1. The order from top to bottom matters, so these are permutations.
2. Add the cases by the addition rule.

Answer: (d) 504

Explanation:
1. Total words \( 6!=720 \).
2. Begin with G: \( 5!=120 \); end with S: \( 120 \); both: \( 4!=24 \).
3. Required \( 720-120-120+24=504 \).

Teacher's Note:
1. Use inclusion-exclusion for 'neither ... nor' conditions.
2. Do not subtract the 'both' case twice.

Answer: (b) 90720

Explanation:
1. After fixing T at both ends the remaining letters are M, A, H, E, M, A, I, C, S (9 letters) in which M occurs twice and A occurs twice.
2. Number of words \( \frac{9!}{2!\,2!}=90720 \).

Teacher's Note:
1. Remove the two T's first, then count arrangements of the rest with repetitions.
2. Divide by the factorial of each repeated letter.

Answer: (b) 2472

Explanation:
1. Let the team have \( b \) bowlers (3 or 4), \( w \) keepers (1 or 2) and the rest from the other 10 players.
2. Cases: \( (3,1): 4\cdot2\cdot\binom{10}{7}=960 \); \( (3,2): 4\cdot1\cdot\binom{10}{6}=840 \); \( (4,1): 1\cdot2\cdot\binom{10}{6}=420 \); \( (4,2): \binom{10}{5}=252 \).
3. Total \( 960+840+420+252=2472 \).

Teacher's Note:
1. Split by the number of bowlers and keepers.
2. The remaining players are chosen from the 10 who are neither bowlers nor keepers.

Answer: (b) 21

Explanation:
1. The 5 must go to the lower deck and the 8 must go to the upper deck.
2. The lower deck needs 2 more and the upper deck needs 5 more from the 7 remaining passengers.
3. Number of ways \( \binom{7}{5}=21 \).

Teacher's Note:
1. Place the forced passengers first.
2. Then choose for the free passengers only.

Answer: (a) 69760

Explanation:
1. All 5-letter words: \( 10^5=100000 \).
2. Words with no letter repeated: \( 10\cdot9\cdot8\cdot7\cdot6=30240 \).
3. Required \( 100000-30240=69760 \).

Teacher's Note:
1. 'At least one repeated' is the complement of 'none repeated'.
2. Repetition is allowed in the total count.

Answer: (d) \( \frac{10!}{3} \)

Explanation:
1. Only the relative order of the three speakers S1, S2, S3 matters; the other seven are free.
2. Of the \( 3!=6 \) possible orders of these three, S2 is last in exactly 2 (S1 S3 S2 and S3 S1 S2).
3. So the number of ways is \( 10!\times\frac26=\frac{10!}{3} \).

Teacher's Note:
1. Only the relative order of three speakers matters.
2. The speaker S2 must be last among the three, which has probability 1/3.

Answer: (c) \( 16\times1111\times3! \)

Explanation:
1. Each digit appears in each place \( 3! \) times.
2. Sum of digits \( 1+3+5+7=16 \), so the sum is \( 3!\times16\times(1000+100+10+1) \).
3. This equals \( 16\times1111\times3!=106656 \).

Teacher's Note:
1. Each digit occupies each position equally often.
2. Sum = (digit sum) x (place values) x (arrangements of the others).

Answer: (d) 60

Explanation:
1. There are 4 even positions (2nd, 4th, 6th, 8th) and 4 odd digits 3, 3, 5, 5: \( \frac{4!}{2!2!}=6 \) ways.
2. The digits 2, 2, 8, 8, 8 fill the 5 odd positions: \( \frac{5!}{2!3!}=10 \) ways.
3. Total \( 6\times10=60 \).

Teacher's Note:
1. Fill the restricted positions first.
2. Identical digits reduce the count by factorials.

Answer: (b) 64

Explanation:
1. Total ways \( \binom93=84 \).
2. Draws with no black ball: \( \binom63=20 \).
3. Required \( 84-20=64 \).

Teacher's Note:
1. 'At least one' means total minus 'none'.
2. Treat balls of the same colour as distinct, as stated.

Answer: (c) 2880

Explanation:
1. Twin prime pairs among 1 to 8: (3, 5) and (5, 7).
2. The top two cards can be a pair in either order: \( 2\times2=4 \) ways.
3. The remaining 6 cards can be arranged in \( 6!=720 \) ways, giving \( 4\times720=2880 \).

Teacher's Note:
1. Twin primes differ by 2.
2. Each pair can occur in 2 orders at the top.

Answer: (a) 2133120

Explanation:
1. Each digit appears in each place \( 4^3=64 \) times.
2. Sum of digits \( 6+7+8+9=30 \), so the total is \( 64\times30\times1111=2133120 \).

Teacher's Note:
1. With repetition each digit appears equally often in every place.
2. Multiply by the place-value sum 1111.

Answer: (a) 2268000

Explanation:
1. EXAMINATION has A×2, I×2, N×2 and the letters E, X, M, T, O.
2. Words starting with A: \( \frac{10!}{2!2!}=907200 \); with E: \( \frac{10!}{2!2!2!}=453600 \); with I: \( \frac{10!}{2!2!}=907200 \).
3. Total before M \( 907200+453600+907200=2268000 \).

Teacher's Note:
1. Only the letters A, E, I come before M alphabetically.
2. Count the arrangements of the remaining 10 letters in each case.

Answer: (c) \( \frac{9!}{3!\,(2!)^3} \)

Explanation:
1. If each elder child gets \( n \) toys the youngest gets \( n+1 \) and \( 3n+n+1=9 \), so \( n=2 \): the shares are 2, 2, 2, 3.
2. Number of ways \( \frac{9!}{2!\,2!\,2!\,3!}=7560 \).

Teacher's Note:
1. Form the equation for the equal share first.
2. Children are distinct, so the shares go to specific children.

Answer: (d) \( \frac{10!}{4} \)

Explanation:
1. Divide 9 people into groups of 2, 3 and 4: \( \frac{9!}{2!3!4!} \) ways.
2. Assign the groups to different floors out of 10: \( {}^{10}P_3=\frac{10!}{7!} \).
3. Total \( \frac{9!}{2!3!4!}\times\frac{10!}{7!}=907200=\frac{10!}{4} \).

Teacher's Note:
1. The groups have different sizes, so they are distinguishable.
2. Choose and order the floors with a permutation.

Answer: (c) 2160

Explanation:
1. Treat the 3 white roses as one block: with the 6 red roses there are 7 objects, giving \( \frac{6!}{2}=360 \) garlands.
2. The white roses can be arranged in \( 3!=6 \) ways.
3. Total \( 360\times6=2160 \).

Teacher's Note:
1. Garland: divide the circular permutation (n - 1)! by 2.
2. Multiply by the internal arrangement of the block.

Answer: (a) \( 3!\cdot4! \)

Explanation:
1. Fix the 4 boys in a circle: \( (4-1)!=3! \) ways.
2. The 4 girls then take the 4 gaps between boys: \( 4! \) ways.
3. Total \( 3!\cdot4!=144 \).

Teacher's Note:
1. Seat one type around the circle first.
2. The other type fills the gaps in any order.

Answer: (b) \( 2\cdot(3!)^3\cdot4! \)

Explanation:
1. Arrange the 4 nationality groups round the circle: \( (4-1)!=3!=6 \).
2. Arrange the people inside each group: \( 2!\cdot3!\cdot3!\cdot4! \).
3. Total \( 3!\cdot2!\cdot3!\cdot3!\cdot4!=2\cdot(3!)^3\cdot4!=10368 \).

Teacher's Note:
1. Treat each nationality as a block, arrange the blocks, then the members.
2. Circular arrangement of 4 blocks gives 3!.

Answer: (a) 12

Explanation:
1. Try numbers: \( 360=2^3\cdot3^2\cdot5 \) has \( 4\cdot3\cdot2=24 \) divisors and no smaller number does (for example 420 and 480 are larger).
2. 360 is not a perfect square, so the number of ways to write it as a product of two factors is \( \frac{24}{2}=12 \).

Teacher's Note:
1. Smaller primes with larger exponents give smaller numbers for a given divisor count.
2. Divide the divisor count by 2 for a non-square.

Answer: (b) 16

Explanation:
1. \( 27720=2^3\cdot3^2\cdot5\cdot7\cdot11 \) has 5 distinct primes.
2. Number of ways \( 2^{5-1}=16 \).

Teacher's Note:
1. Each prime power must go wholly into one of the two factors.
2. Unordered pairs: 2^(k-1) for k distinct primes.

Answer: (b) 953

Explanation:
1. All selections: \( 6\times5\times4\times2^3=960 \) (including the empty selection).
2. Remove the empty selection (1) and the selections of exactly one fruit: 1 mango, 1 apple, 1 banana and 3 different fruits, i.e. 6.
3. Required \( 960-1-6=953 \).

Teacher's Note:
1. Identical fruits: choose a count for each kind.
2. At least two fruits excludes the empty and single-fruit selections.

Answer: (a) 27

Explanation:
1. Number of solutions of \( x+y+z=11 \) with \( 1\le x,y,z\le6 \).
2. Without the upper limit: \( \binom{10}{2}=45 \); subtract the cases in which one variable is 7 or more: \( 3\times\binom42=18 \).
3. Required \( 45-18=27 \).

Teacher's Note:
1. Subtract the solutions in which a die shows more than 6.
2. At most one die can exceed 6 here.