JEE Mathematics Permutations and Combinations MCQs Set 20

Answer: (a) 180

Explanation:
1. The 9 places include 4 even places (2nd, 4th, 6th, 8th). The three odd digits 1, 1, 3 must go into these 4 places.
2. Ways to place the odd digits: \( {}^4C_3\times\frac{3!}{2!}=4\times3=12 \).
3. The remaining 6 places take 2, 2, 2, 2, 4, 4 in \( \frac{6!}{4!\,2!}=15 \) ways. Total \( =12\times15=180 \). Answer: (a)

Teacher's Note:
1. Choose which 3 of the 4 even places get odd digits.
2. Divide by factorials for repeated digits.

Answer: (c) \( m=n=78 \)

Explanation:
1. At least 6 males: \( {}^8C_6{}^5C_5+{}^8C_7{}^5C_4+{}^8C_8{}^5C_3=28+40+10=78 \).
2. At least 3 females: \( {}^5C_3{}^8C_8+{}^5C_4{}^8C_7+{}^5C_5{}^8C_6=10+40+28=78 \).
3. So \( m=n=78 \). Answer: (c)

Teacher's Note:
1. Note that 11 members with at least 6 males means at most 5 females.
2. Add the cases for each allowed composition.

Answer: (a) 60

Explanation:
1. Digit sum \( =24 \). Divisibility by 11 needs (sum of odd-place digits) \( - \) (sum of even-place digits) to be 0 or a multiple of 11; with total 24 the only possibility is each sum \( =12 \).
2. Sets with sum 12: \( \{0,5,7\} \) and \( \{1,2,9\} \). If the odd places have \( \{0,5,7\} \) the leading digit cannot be 0: \( 2\cdot2!\times3!=24 \) numbers.
3. If the odd places have \( \{1,2,9\} \): \( 3!\times3!=36 \) numbers. Total \( =24+36=60 \). Answer: (a)

Teacher's Note:
1. Use the divisibility rule of 11 with alternate digit sums.
2. Take care that the first digit is non-zero.

Answer: (c) \( 2^{20} \)

Explanation:
1. If \( r \) identical objects are taken, the other \( 10-r \) come from 21 distinct ones: total \( x=\sum_{r=0}^{10}{}^{21}C_{10-r}={}^{21}C_0+{}^{21}C_1+\cdots+{}^{21}C_{10} \).
2. Since \( {}^{21}C_{k}={}^{21}C_{21-k} \), the sum \( {}^{21}C_{11}+\cdots+{}^{21}C_{21} \) also equals \( x \).
3. Adding, \( 2x=2^{21} \), so \( x=2^{20} \). Answer: (c)

Teacher's Note:
1. Split by the number of identical objects chosen.
2. Use the symmetry of binomial coefficients.

Answer: (c) 25

Explanation:
1. Teams: 1 boy 2 girls or 2 boys 1 girl: \( 5\cdot\frac{n(n-1)}2+10n=1750 \).
2. \( n(n-1)+4n=700\Rightarrow n^2+3n-700=0 \).
3. \( (n+28)(n-25)=0\Rightarrow n=25 \). Answer: (c)

Teacher's Note:
1. List the two allowed compositions of the team.
2. Form and solve a quadratic in \( n \).

Answer: (a) 374

Explanation:
1. 1-digit: 4; 2-digit: \( 4\times5=20 \); 3-digit: \( 4\times5\times5=100 \).
2. 4-digit numbers below 7000: first digit 1 or 3 (2 ways), others 5 each: \( 2\times5^3=250 \).
3. Total \( =4+20+100+250=374 \). Answer: (a)

Teacher's Note:
1. Count numbers of each length separately.
2. Zero cannot be the first digit, and 7 or 9 cannot start a 4-digit number below 7000.

Answer: (b) 120

Explanation:
1. Any 3 different labels from 1 to 10 can be assigned in exactly one increasing order to the three boxes.
2. So the number of ways is \( {}^{10}C_3=120 \). Answer: (b)

Teacher's Note:
1. An increasing order is determined by the chosen set of labels.
2. Count the 3-element subsets.

Answer: (a) at least 1000

Explanation:
1. Select 4 novels: \( {}^6C_4=15 \) ways; select 1 dictionary: 3 ways.
2. The dictionary is fixed in the middle, and the 4 novels are arranged in the remaining places in \( 4!=24 \) ways.
3. Total \( =15\times3\times24=1080 \), which is at least 1000. Answer: (a)

Teacher's Note:
1. Select first, then arrange.
2. The middle place is fixed for the dictionary.

Answer: (a) 485

Explanation:
1. Let \( X \)'s three guests include \( i \) ladies and \( 3-i \) men. Then \( Y \)'s three guests must include \( 3-i \) ladies and \( i \) men, to make 3 ladies and 3 men in all.
2. The number of ways is \( \left[{}^4C_i\,{}^3C_{3-i}\right]\left[{}^3C_{3-i}\,{}^4C_i\right] \): for \( i=0,1,2,3 \) this is \( 1\cdot1,\ 12\cdot12,\ 18\cdot18,\ 4\cdot4 \).
3. Total \( =1+144+324+16=485 \). Answer: (a)

Teacher's Note:
1. Let the number of ladies from X's friends be \( i \).
2. Then Y supplies the remaining ladies and the men are fixed automatically.

Answer: (a) 219

Explanation:
1. \( n(A\times B)=8 \), so there are \( 2^8=256 \) subsets in all.
2. Subsets with fewer than 3 elements: \( {}^8C_0+{}^8C_1+{}^8C_2=1+8+28=37 \).
3. Required \( =256-37=219 \). Answer: (a)

Teacher's Note:
1. Count the complement: subsets with 0, 1 or 2 elements.
2. \( n(A\times B)=n(A)\,n(B) \).

Answer: (b) 192

Explanation:
1. 4-digit numbers above 6000 must start with 6, 7 or 8: \( 3\times4\times3\times2=72 \).
2. All 5-digit numbers exceed 6000: \( 5!=120 \).
3. Total \( =72+120=192 \). Answer: (b)

Teacher's Note:
1. Treat 4-digit and 5-digit numbers separately.
2. For 4 digits, the first digit must be 6, 7 or 8.

Answer: (b) 5

Explanation:
1. \( T_n={}^nC_3 \), so \( T_{n+1}-T_n={}^{n+1}C_3-{}^nC_3={}^nC_2 \).
2. \( {}^nC_2=10\Rightarrow\frac{n(n-1)}2=10\Rightarrow n=5 \). Answer: (b)

Teacher's Note:
1. Use Pascal's rule \( {}^{n+1}C_3={}^nC_3+{}^nC_2 \).
2. Solve \( n(n-1)=20 \).

Answer: (d) 879

Explanation:
1. Choose 0 to 10 white balls (11 ways), 0 to 9 green (10 ways), 0 to 7 black (8 ways): \( 11\times10\times8=880 \).
2. This includes the choice of no ball at all, which must be removed: \( 880-1=879 \). Answer: (d)

Teacher's Note:
1. Balls of the same colour are identical, so only the count of each colour matters.
2. Subtract the empty selection.

Answer: (b) \( N\le100 \)

Explanation:
1. Triangles from any 3 points: \( {}^{10}C_3=120 \).
2. Remove the degenerate triples from the 6 collinear points: \( {}^6C_3=20 \).
3. \( N=100 \), so \( N\le100 \). Answer: (b)

Teacher's Note:
1. Three collinear points do not form a triangle.
2. Subtract the collinear triples from the total.

Answer: (d) A is false, but R is true.

Explanation:
1. Let \( x_i \) be the number of ice-creams of type \( i \): \( x_1+\cdots+x_5=6 \), non-negative integers. The number of solutions is \( {}^{6+5-1}C_{5-1}={}^{10}C_4=210 \), not \( {}^{10}C_5=252 \). So the Assertion is false.
2. Arranging 6 A's and 4 B's in a row: \( \frac{10!}{6!\,4!}=210 \), which equals the number of ways to buy the ice-creams. The Reason is true. Answer: (d)

Teacher's Note:
1. Count non-negative integer solutions using stars and bars.
2. Arrangements of identical letters use \( \frac{n!}{p!\,q!} \).

Answer: (c) \( \frac{12!}{(4!)^3} \)

Explanation:
1. Since \( A,B,C \) are named sets, choose 4 elements for \( A \), then 4 of the remaining 8 for \( B \), and the last 4 form \( C \).
2. \( {}^{12}C_4\times{}^8C_4\times{}^4C_4=\frac{12!}{8!\,4!}\cdot\frac{8!}{4!\,4!}\cdot1=\frac{12!}{(4!)^3} \). Answer: (c)

Teacher's Note:
1. Distinct labels A, B, C mean no division by 3! is needed.
2. Multiply the successive selections.

Answer: (b) 385

Explanation:
1. He votes for 1, 2, 3 or 4 candidates.
2. \( {}^{10}C_1+{}^{10}C_2+{}^{10}C_3+{}^{10}C_4=10+45+120+210=385 \). Answer: (b)

Teacher's Note:
1. Add the cases of voting for 1, 2, 3 and 4 candidates.
2. Order of choosing does not matter.

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

Explanation:
1. For \( {}^nP_r \) we need \( n\ge r\ge0 \): \( 7-x\ge x-3\ge0 \), so \( 3\le x\le5 \), and \( x \) is an integer.
2. \( f(3)={}^4P_0=1 \), \( f(4)={}^3P_1=3 \), \( f(5)={}^2P_2=2 \).
3. Range \( =\{1,2,3\} \). Answer: (a)

Teacher's Note:
1. Find the domain from the conditions on \( n \) and \( r \) in \( {}^nP_r \).
2. Evaluate the function at each domain value.

Answer: (c) 360

Explanation:
1. All 6 letters can be arranged in \( 6!=720 \) ways.
2. The two vowels A and E can appear in 2 orders and exactly one of them is alphabetical, so divide by \( 2! \).
3. \( \frac{720}{2}=360 \). Answer: (c)

Teacher's Note:
1. Half of all arrangements have A before E.
2. Divide by the number of orders of the vowels.

Answer: (b) 196

Explanation:
1. Exactly 4 from the first five: \( {}^5C_4\times{}^8C_6=5\times28=140 \).
2. All 5 from the first five: \( {}^5C_5\times{}^8C_5=56 \).
3. Total \( =140+56=196 \). Answer: (b)

Teacher's Note:
1. Split into the cases 4 and 5 questions from the first five.
2. The rest come from the last 8.

Answer: (a) \( 6!\times5! \)

Explanation:
1. Seat the 6 men at the round table first: \( (6-1)!=5! \) ways. This leaves 6 gaps between men.
2. Place the 5 women in 5 of the 6 gaps: \( {}^6P_5=6! \) ways.
3. Total \( =5!\times6! \). Answer: (a)

Teacher's Note:
1. Fix the men first in circular order.
2. Women then go into separate gaps.

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

Explanation:
1. Write \( 2\,{}^nC_r={}^nC_r+{}^nC_r \) and group: \( ({}^nC_{r+1}+{}^nC_r)+({}^nC_{r-1}+{}^nC_r) \).
2. By Pascal's rule \( {}^nC_{r+1}+{}^nC_r={}^{n+1}C_{r+1} \) and \( {}^nC_{r-1}+{}^nC_r={}^{n+1}C_r \).
3. \( {}^{n+1}C_{r+1}+{}^{n+1}C_r={}^{n+2}C_{r+1} \). Answer: (b)

Teacher's Note:
1. Pair each neighbouring term with one copy of \( {}^nC_r \).
2. Apply Pascal's rule twice.