JEE Mathematics Permutation and Combination MCQs Set 10

Answer: (b) 12

Explanation:
1. A polygon with n sides has \( \dfrac{n(n-3)}2 \) diagonals.
2. \( n(n-3)=108 \), so \( n^2-3n-108=0 \), i.e. \( (n-12)(n+9)=0 \).
3. Therefore \( n=12 \) (the negative root is rejected).

Teacher's Note:
1. Diagonals = nC2 minus n sides.

Answer: (a) \( \sum_{k=1}^{n-1}k \)

Explanation:
1. Every pair of lines meets in exactly one point, and no point is shared by three lines.
2. The number of points is \( {}^nC_2=\dfrac{n(n-1)}2 \).
3. This equals \( 1+2+\dots+(n-1)=\sum_{k=1}^{n-1}k \).

Teacher's Note:
1. Count pairs of lines.

Answer: (c) 5

Explanation:
1. Triangles \( ={}^{10}C_3-{}^nC_3=120-{}^nC_3 \).
2. \( {}^nC_3=10 \) gives \( n=5 \).
3. Check: \( {}^5C_3=10 \) and \( 120-10=110 \).

Teacher's Note:
1. Subtract the degenerate triples of collinear points.

Answer: (c) \( p^2(4p-3) \)

Explanation:
1. Total triples of the 3p points: \( {}^{3p}C_3=\dfrac{p(3p-1)(3p-2)}2 \).
2. Subtract the collinear triples on the three lines: \( 3\cdot{}^pC_3=\dfrac{p(p-1)(p-2)}2 \).
3. The difference is \( \dfrac p2(8p^2-6p)=p^2(4p-3) \).

Teacher's Note:
1. Maximum triangles when no other three points are collinear.

Answer: (d) none of these

Explanation:
1. Choose b black (1 to 3), w white (1 to 4) and r red (1 or 2) with \( b+w+r\le6 \): the number is \( \sum{}^3C_b\,{}^4C_w\,{}^2C_r \).
2. Evaluating over all allowed triples (b, w, r) gives 270.
3. The options give \( 1008,\ 1536,\ 1512 \), none of which is 270.

Teacher's Note:
1. List the allowed colour counts and add the products of combinations.

Answer: (a) 30

Explanation:
1. Using only 1 and 2 there are \( 2^5=32 \) five-digit numbers.
2. The numbers in which all digits are the same are 11111 and 22222, i.e. 2 numbers.
3. At least one different digit: \( 32-2=30 \).

Teacher's Note:
1. Use the complement: all numbers minus those with identical digits.

Answer: (c) 6

Explanation:
1. Let there be n contestants, so a voter can vote for at most \( n-1 \) candidates.
2. The number of ways of voting for 1 to \( n-1 \) candidates is \( 2^n-2 \).
3. \( 2^n-2=62 \) gives \( n=6 \).

Teacher's Note:
1. Sum the selections of 1 to n - 1 candidates.

Answer: (c) \( 6^n-3^n \)

Explanation:
1. Total outcomes of n dice: \( 6^n \).
2. Outcomes with no odd number (all even): \( 3^n \).
3. At least one odd: \( 6^n-3^n \).

Teacher's Note:
1. Use the complement of all-even outcomes.

Answer: (b) \( 2\times3^5 \)

Explanation:
1. The first digit cannot be 0, so it has 2 choices.
2. Each of the other 5 places has 3 choices.
3. The total is \( 2\times3^5 \).

Teacher's Note:
1. A number cannot start with 0.

Answer: (b) \( n^3-{}^nP_3 \)

Explanation:
1. The total number of triplets is \( n^3 \).
2. Triplets with all coordinates different number \( {}^nP_3 \).
3. At least two equal: \( n^3-{}^nP_3 \).

Teacher's Note:
1. Use the complement of all-distinct triplets.

Answer: (d) none of these

Explanation:
1. The odd digits are 1, 3, 5, 7, 9, so all 5 appear in 6 places and exactly one digit is repeated.
2. Choose the repeated digit in 5 ways; arrange the 6 digits with that digit twice in \( \dfrac{6!}{2!} \) ways.
3. The total is \( 5\cdot\dfrac{6!}{2}=1800=\tfrac52(6!) \), which is not among the options.

Teacher's Note:
1. One of the five digits must appear twice.

Answer: (a) 4

Explanation:
1. \( 240=2^4\cdot3\cdot5 \). A divisor of the form \( 4n+2 \) is twice an odd number.
2. So it has exactly one factor of 2 and an odd part from \( 3\cdot5 \): the odd divisors are 1, 3, 5, 15.
3. The divisors are 2, 6, 10, 30: four in all.

Teacher's Note:
1. A number congruent to 2 mod 4 has exactly one factor 2.

Answer: (a) \( 60(5!) \)

Explanation:
1. If A is in position i, B is in position \( i+3 \) or \( i-3 \): the pairs of positions are (1,4), (2,5), (3,6), (4,7), (5,8), i.e. 5 pairs.
2. A and B can swap in 2 ways, so \( 10 \) ways to place them; the two persons between and the rest: choose the other 6 people in the remaining 6 places in \( 6! \) ways.
3. Total \( 10\cdot6!=7200=60\cdot5! \).

Teacher's Note:
1. Place A and B first, then arrange the others.

Answer: (a) 66

Explanation:
1. Gentlemen in majority means 3 gentlemen and 2 ladies, or 4 gentlemen and 1 lady.
2. 3G + 2L: \( {}^4C_3\cdot{}^6C_2=4\cdot15=60 \). 4G + 1L: \( {}^4C_4\cdot{}^6C_1=6 \).
3. Total \( 60+6=66 \).

Teacher's Note:
1. List the possible compositions of the majority.

Answer: (c) \( {}^{20}C_{10} \)

Explanation:
1. All students solve the same number k of questions, and their sets of solved questions are all different.
2. The number of different combinations is \( {}^{20}C_k \).
3. The maximum is attained at \( k=10 \), giving \( {}^{20}C_{10} \).

Teacher's Note:
1. nCk is largest for k equal to n/2.

Answer: (c) 120

Explanation:
1. \( 24=2^3\cdot3 \). Positive solutions: distribute the three 2's in \( {}^5C_2=10 \) ways and the 3 in \( {}^3C_1=3 \) ways, giving 30.
2. Signs: an even number of the three variables is negative: \( 1+{}^3C_2=4 \) sign patterns.
3. Total \( 30\times4=120 \).

Teacher's Note:
1. Count positive solutions first and then the sign patterns that keep the product positive.

Answer: (a) 576

Explanation:
1. ARTICLE has 7 letters: consonants R, T, C, L (4) and vowels A, I, E (3). The even places are 2, 4 and 6.
2. Fill the three even places with consonants: \( {}^4P_3=24 \) ways.
3. Arrange the remaining 4 letters in the other 4 places: \( 4!=24 \). Total \( 24\times24=576 \).

Teacher's Note:
1. Fill the restricted places first.

Answer: (a) 1440

Explanation:
1. Treat the couple as one unit: there are \( 6+1=7 \) units around the table, arranged in \( (7-1)!=720 \) ways.
2. The two members of the couple can swap places in 2 ways.
3. Total \( 720\times2=1440 \).

Teacher's Note:
1. Tie the couple together, then multiply by the 2 internal orders.

Answer: (c) 4!

Explanation:
1. There are 6 kinds. From mangoes choose 1 to 4: 4 ways; apples 1 to 3: 3 ways; oranges 1 to 2: 2 ways.
2. Each of the three other varieties is taken once: 1 way each.
3. Total \( 4\cdot3\cdot2\cdot1\cdot1\cdot1=24=4! \).

Teacher's Note:
1. Multiply the number of choices for each kind.