JEE Mathematics Permutations and Combinations MCQs Set 25

Answer: (a) 180

Explanation:
1. There are 4 even places and 5 odd places among the nine positions. The three odd digits 1, 1, 3 must go in even places.
2. Choose 3 of the 4 even places and arrange 1, 1, 3: \( {}^4C_3\times\dfrac{3!}{2!}=12 \).
3. Fill the remaining 6 places with 2, 2, 2, 2, 4, 4: \( \dfrac{6!}{4!\,2!}=15 \).
4. Total \( =12\times15=180 \).

Teacher's Note:
1. Place the restricted digits first.
2. Then arrange the remaining digits, dividing by the factorials of repeated digits.

Answer: (a) 310

Explanation:
1. Starting with 5: \( 6^3=216 \) numbers.
2. Starting with 4 and second digit 4 or 5: \( 2\times6\times6=72 \).
3. Starting with 43 and third digit 3, 4 or 5: \( 3\times6=18 \); starting with 432 and last digit 2 to 5: 4.
4. Total \( =216+72+18+4=310 \).

Teacher's Note:
1. Count by the leading digit and compare place by place with 4321.

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

Explanation:
1. At least 6 males (committee of 11 has at most 5 females): \( {}^8C_6\,{}^5C_5+{}^8C_7\,{}^5C_4+{}^8C_8\,{}^5C_3=28+40+10=78 \).
2. At least 3 females: \( {}^5C_3\,{}^8C_8+{}^5C_4\,{}^8C_7+{}^5C_5\,{}^8C_6=10+40+28=78 \).
3. So \( m=n=78 \).

Teacher's Note:
1. At least 6 males among 11 is the same event as at least 3 females in this committee, so m and n are equal.

Answer: (a) 60

Explanation:
1. Sum of the digits is 24. Let the number be abcdef; divisibility by 11 requires \( (a+c+e)-(b+d+f) \) to be 0 or a multiple of 11.
2. Only \( a+c+e=b+d+f=12 \) is possible. The splits are \( \{0,5,7\},\{1,2,9\} \) and \( \{1,2,9\},\{0,5,7\} \).
3. Case 1 (0, 5, 7 at odd places): first place cannot be 0, so \( 2\times2!\times3!=24 \). Case 2: \( 3!\times3!=36 \).
4. Total \( =24+36=60 \).

Teacher's Note:
1. Use the divisibility rule for 11 on alternate digit sums, and remember a six-digit number cannot start with zero.

Answer: (a) 170

Explanation:
1. Each pair of pillars could be joined: \( {}^{20}C_2=190 \) pairs.
2. Remove the 20 adjacent pairs (sides): \( 190-20=170 \).

Teacher's Note:
1. This is the count of diagonals of a 20-sided polygon: n(n minus 3) over 2.

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

Explanation:
1. Choose k identical objects (one way) and \( 10-k \) distinct ones: total \( \sum_{k=0}^{10}{}^{21}C_{10-k}={}^{21}C_0+{}^{21}C_1+\dots+{}^{21}C_{10} \).
2. Since \( {}^{21}C_r={}^{21}C_{21-r} \), this is half of \( 2^{21} \).
3. So the number of ways is \( 2^{20} \).

Teacher's Note:
1. Because 21 is odd, the binomial coefficients up to 10 are exactly half of the full sum 2 to the power 21.

Answer: (a) 25

Explanation:
1. Teams with 1 boy and 2 girls plus 2 boys and 1 girl: \( 5\,{}^nC_2+10\,n=1750 \).
2. \( \dfrac{5n(n-1)}{2}+10n=1750 \), so \( n^2-n+4n=700 \), \( n^2+3n-700=0 \).
3. \( n=25 \).

Teacher's Note:
1. Split the mixed teams into the two possible compositions and solve the resulting quadratic.

Answer: (a) 374

Explanation:
1. One-digit: 4 (1, 3, 7, 9). Two-digit: \( 4\times5=20 \). Three-digit: \( 4\times5\times5=100 \).
2. Four-digit less than 7000: first digit 1 or 3 (2 choices): \( 2\times5^3=250 \).
3. Total \( =4+20+100+250=374 \).

Teacher's Note:
1. Count by number of digits.
2. For four digits, the first digit must be below 7 and non-zero.

Answer: (b) 120

Explanation:
1. Any three distinct labels can be placed in increasing order in exactly one way.
2. The number of ways is \( {}^{10}C_3=120 \).

Teacher's Note:
1. An increasing assignment is determined by choosing which three distinct labels appear.

Answer: (a) 12

Explanation:
1. Games among men \( =2\,{}^mC_2=m(m-1) \).
2. Games between men and women \( =2\times2m=4m \).
3. \( m(m-1)-4m=84 \), so \( m^2-5m-84=0 \) and \( m=12 \).

Teacher's Note:
1. Count each pairing twice because every pair plays two games.

Answer: (b) 1080

Explanation:
1. Choose 4 novels: \( {}^6C_4=15 \). Choose 1 dictionary: \( {}^3C_1=3 \).
2. Dictionary fixed in the middle; arrange the 4 novels in the other positions: \( 4!=24 \).
3. \( 15\times3\times24=1080 \).

Teacher's Note:
1. Select first, then arrange.
2. The dictionary has no freedom of position.

Answer: (c) 485

Explanation:
1. Let X invite k ladies and \( 3-k \) men; then Y must invite \( 3-k \) ladies and k men.
2. \( \sum_{k=0}^{3}{}^4C_k\,{}^3C_{3-k}\,{}^3C_{3-k}\,{}^4C_k=1+144+324+16 \).
3. Total \( =485 \).

Teacher's Note:
1. Condition on how many ladies X brings; Y's invitation is then forced.

Answer: (b) 58th

Explanation:
1. Letters in order: A, L, L, M, S. Words beginning with A: \( \dfrac{4!}{2!}=12 \); with L: \( 4!=24 \); with M: 12. Total 48.
2. Beginning with SA: \( \dfrac{3!}{2!}=3 \); with SL: \( 3!=6 \).
3. SMALL is the first word beginning with SM: position \( 48+3+6+1=58 \).

Teacher's Note:
1. Count the words that precede it letter by letter.
2. Divide by 2 factorial only when the remaining letters include both L's.

Answer: (a) 219

Explanation:
1. \( n(A\times B)=8 \), so the number of subsets is \( 2^8=256 \).
2. Subtract subsets with 0, 1 or 2 elements: \( 1+8+28=37 \).
3. \( 256-37=219 \).

Teacher's Note:
1. Use the complement: all subsets minus those with fewer than three elements.

Answer: (a) 192

Explanation:
1. Four-digit numbers greater than 6000: first digit from 6, 7, 8 (3 ways), then \( 4\times3\times2 \): \( 3\times24=72 \).
2. Five-digit numbers: \( 5!=120 \).
3. Total \( =72+120=192 \).

Teacher's Note:
1. Treat four-digit and five-digit numbers separately.

Answer: (d) 5

Explanation:
1. \( T_n={}^nC_3 \).
2. \( {}^{n+1}C_3-{}^nC_3={}^nC_2=10 \).
3. \( n(n-1)=20 \), so \( n=5 \).

Teacher's Note:
1. Pascal's rule turns the difference into a single binomial coefficient.

Answer: (b) 879

Explanation:
1. Choices of white balls: 0 to 10 (11 ways); green: 10 ways; black: 8 ways.
2. Total selections \( =11\times10\times8=880 \), including the empty selection.
3. Excluding it: 879.

Teacher's Note:
1. For identical items of each colour, count the choices of how many of each colour and subtract the empty selection.

Answer: (b) \( N\leq100 \)

Explanation:
1. Triangles from any 3 points: \( {}^{10}C_3=120 \).
2. Three collinear points give no triangle: subtract \( {}^6C_3=20 \).
3. \( N=100 \), so \( N\leq100 \).

Teacher's Note:
1. Subtract the choices of three collinear points, since they do not form a triangle.

Answer: (a) 210

Explanation:
1. Let \( x_1+\dots+x_5=6 \) with \( x_i\ge0 \).
2. Number of solutions \( ={}^{6+5-1}C_{5-1}={}^{10}C_4=210 \).
3. This also equals the number of arrangements of 6 identical A's and 4 identical B's in a row.

Teacher's Note:
1. Selecting with repetition from 5 types is a stars and bars count.