JEE Mathematics Permutation and Combination MCQs Set 16

Answer: (a) \( 86 \)

Explanation:
1. Let the equal sides be \( b \) and the base \( a \): \( a+2b=2008 \), so \( a \) is even and \( a \lt 2b \) (triangle inequality).
2. Obtuse at the apex: \( a^2 \gt 2b^2 \). With \( a=2008-2b \): \( (2008-2b)^2 \gt 2b^2 \) gives \( b \lt \frac{2008}{2+\sqrt2}\approx588.1 \), and \( a \lt 2b \) gives \( b \gt 502 \).
3. The integers \( b=503,\dots,588 \) number \( 86 \).

Teacher's Note:
1. Impose the triangle inequality for a lower bound on \( b \) and the obtuse condition \( a^2 \gt 2b^2 \) for an upper bound.

Answer: (a) \( 5 \)

Explanation:
1. Blue segments (adjacent pairs): \( n \). All segments: \( \binom n2 \), so red \( =\binom n2-n \).
2. Equal numbers: \( \binom n2-n=n \), so \( \frac{n(n-1)}{2}=2n \) and \( n-1=4 \).
3. \( n=5 \).

Teacher's Note:
1. The adjacent pairs form the sides of the polygon, which number \( n \).

Answer: (d) \( 141 \)

Explanation:
1. We need the coefficient of \( x^0 \) in \( \left(x^{-1}+1+x\right)^6 \), i.e. the central trinomial coefficient.
2. Count by the number \( k \) of \( +1 \) cards (equal to the number of \( -1 \) cards): \( \sum_{k=0}^{3}\frac{6!}{k!\,k!\,(6-2k)!}=1+30+90+20=141 \).

Teacher's Note:
1. Equal numbers of \( +1 \) and \( -1 \) cards are needed; count the arrangements for each possible number of such pairs.

Answer: (c) \( 420 \)

Explanation:
1. Total ways to choose \( 4 \) points: \( \binom{12}{4}=495 \).
2. Subtract choices that cannot form a quadrilateral: \( 4 \) collinear points \( \binom54=5 \), and \( 3 \) collinear plus one other: \( \binom53\cdot7=70 \).
3. \( 495-5-70=420 \).

Teacher's Note:
1. Collinear triples make the fourth point degenerate. Count the selections with 3 or 4 collinear points and remove them.

Answer: (d) \( 41 \)

Explanation:
1. Each element goes to the first subset, the second subset or neither: \( 3^4=81 \) ordered pairs \( (A,B) \) of disjoint subsets.
2. The ordered pair has \( A=B \) only when both are empty (1 case). Unordered pairs: \( \frac{81-1}{2}+1=41 \).

Teacher's Note:
1. Count ordered pairs first and halve, adjusting for the single symmetric case \( A=B=\emptyset \).

Answer: (b) \( \frac{n(n+1)}{2} \)

Explanation:
1. If the boxes hold \( a,b,c \) with \( a+b+c=2n+1 \), then \( a+b \gt c \) means \( c \lt \frac{2n+1}{2} \), i.e. \( c\le n \). So every box holds at most \( n \).
2. Total solutions: \( \binom{2n+3}{2} \). Subtract those with a box holding \( \ge n+1 \): \( 3\binom{n+2}{2} \) (two such boxes cannot occur together).
3. \( \binom{2n+3}{2}-3\binom{n+2}{2}=\frac{(2n+3)(2n+2)-3(n+2)(n+1)}{2}=\frac{n^2+n}{2} \). Check: \( n=2 \) gives \( 3 \).

Teacher's Note:
1. The triangle-type conditions reduce to each box holding at most \( n \) balls. Use complementary counting with the 'at least \( n+1 \)' cases.

Answer: (c) \( (n+2)\,2^{n-1} \)

Explanation:
1. Choose \( k \) of the \( n \) unlike letters in \( \binom nk \) ways. The remaining \( n-k \) letters are \( a \)'s and \( b \)'s: \( n-k+1 \) ways (number of \( a \)'s from \( 0 \) to \( n-k \)).
2. Total \( =\sum_{k=0}^{n}\binom nk(n-k+1)=2^n+n2^{n-1}=(n+2)2^{n-1} \).
3. Check: \( n=2 \) gives \( 8 \).

Teacher's Note:
1. Treat the \( a \)'s and \( b \)'s as two identical types and the unlike letters as distinct. Use \( \sum k\binom nk=n2^{n-1} \).

Answer: (b) \( 121 \)

Explanation:
1. Equal digits: \( 111,222,\dots,999 \) give \( 9 \) numbers. If two digits are equal and the third differs, no digit can be the average of the other two, so all digits must be distinct, in A.P.: \( \{x-d,x,x+d\} \).
2. Such sets: \( d=1 \): \( x=1..8 \) (\( 8 \) sets); \( d=2 \): \( 6 \) sets; \( d=3 \): \( 4 \) sets; \( d=4 \): \( 2 \) sets; total \( 20 \). Four of them contain \( 0 \) (\( \{0,1,2\},\{0,2,4\},\{0,3,6\},\{0,4,8\} \)) and give \( 6-2=4 \) valid numbers each; the other \( 16 \) sets give \( 6 \) each.
3. Total \( =9+16\times6+4\times4=9+96+16=121 \).

Teacher's Note:
1. A systematic count: write the three digits as an arithmetic progression, count permutations, and exclude a leading zero. The total is \( 121 \).

Answer: (a) \( 2186 \)

Explanation:
1. The elements \( 5,7,8 \) are in both sets. Each of the other \( 7 \) elements is in \( A \) only, \( B \) only or in neither: \( 3^7=2187 \) pairs.
2. The case \( A=B \) occurs only when every other element is in neither set: \( 1 \) pair. Excluding it: \( 2186 \).

Teacher's Note:
1. Assign each remaining element to one of three states and remove the single pair with \( A=B \).

Answer: (a) \( 150 \)

Explanation:
1. Total distributions: \( 3^5=243 \). Subtract those where some person gets nothing: \( 3\cdot2^5-3\cdot1^5=93 \).
2. \( 243-93=150 \) (alternatively \( 3!\,S(5,3)=6\times25 \)).

Teacher's Note:
1. Use inclusion-exclusion on persons who receive no ball. This counts onto functions from 5 balls to 3 persons.

Answer: (d) \( 56 \)

Explanation:
1. Maximum marks \( =12 \), so he needs at least \( 3 \) marks. With \( k \) correct matches: marks \( =2k-(6-k)=3k-6\ge3 \), so \( k\ge3 \).
2. \( k=3 \): \( \binom63\cdot D_3=20\cdot2=40 \). \( k=4 \): \( \binom64\cdot D_2=15 \). \( k=5 \) is impossible. \( k=6 \): \( 1 \).
3. Total \( =40+15+1=56 \).

Teacher's Note:
1. Express the marks in terms of the number of correct matches, then count permutations with exactly \( k \) fixed points.

Answer: (c) \( 53 \)

Explanation:
1. Card \( 1 \) goes to envelope \( 2 \). Consider card \( 2 \): either it goes into envelope \( 1 \) (then the remaining \( 4 \) cards form a derangement: \( D_4=9 \) ways) or it goes elsewhere.
2. If card \( 2 \) does not go to envelope \( 1 \), treat envelope \( 1 \) as the 'forbidden envelope' for card \( 2 \); the remaining five cards and five envelopes then behave like a derangement of \( 5 \): \( D_5=44 \).
3. Total \( =D_4+D_5=9+44=53 \).

Teacher's Note:
1. This is a classical derangement count: \( D_n \) with a fixed first assignment equals \( D_{n-1}+D_{n-2} \) for the remaining objects.

Answer: (c) \( 480 \)

Explanation:
1. \( 2310=2\cdot3\cdot5\cdot7\cdot11 \).
2. Euler's function: \( \varphi(2310)=2310\cdot\frac12\cdot\frac23\cdot\frac45\cdot\frac67\cdot\frac{10}{11}=1\cdot2\cdot4\cdot6\cdot10=480 \).
3. The integer \( 2310 \) itself is not coprime to \( 2310 \), so the count below it is \( 480 \).

Teacher's Note:
1. Use \( \varphi(n)=n\prod\left(1-\frac1p\right) \) over the prime factors of \( n \).

Answer: (a) \( 8 \)

Explanation:
1. \( b_n \) counts valid \( n \)-digit numbers ending in \( 1 \); removing the last \( 1 \) leaves any valid \( (n-1) \)-digit number, so \( b_n=a_{n-1} \).
2. \( a_1=1,a_2=2,a_3=3,a_4=5,a_5=8 \), so \( b_6=a_5=8 \).

Teacher's Note:
1. A number ending in \( 1 \) can be preceded by any valid shorter number. The sequence \( 1,2,3,5,8 \) is Fibonacci.

Answer: (d) \( 377 \)

Explanation:
1. Considering the last term: \( f(n)=f(n-1)+f(n-2) \), with \( f(1)=1 \), \( f(2)=2 \).
2. \( f(3)=3 \), \( f(4)=5 \) (matches the given value), \( f(5)=8 \), \( f(6)=13 \).
3. Continuing the Fibonacci sequence: \( f(7)=21 \), \( f(8)=34 \), \( f(9)=55 \), \( f(10)=89 \), \( f(11)=144 \), \( f(12)=233 \), \( f(13)=377 \). So \( f(f(6))=f(13)=377 \).

Teacher's Note:
1. Condition on whether the last term is \( 1 \) or \( 2 \) to get the Fibonacci recurrence. Then evaluate \( f \) at \( 6 \) and again at the result.

Answer: (b) \( 45360 \)

Explanation:
1. Count five-digit strings (a leading zero allowed first): choose the repeated digit (\( 10 \) ways), choose the other three distinct digits from the remaining \( 9 \) (\( \binom93=84 \) ways) and arrange the five digits with one digit repeated: \( \frac{5!}{2!}=60 \). Total \( =10\times84\times60=50400 \).
2. By symmetry exactly \( \frac1{10} \) of these strings start with \( 0 \) (\( 5040 \) strings), and these are not five-digit numbers.
3. Hence the required count is \( 50400-5040=45360 \).

Teacher's Note:
1. By symmetry exactly \( \frac1{10} \) of all digit strings start with \( 0 \), so the valid count is \( \frac{9}{10} \) of the total.

Answer: (b) \( 729 \)

Explanation:
1. Let the number be \( abcd \) with digits \( 1 \) to \( 9 \). Divisible by \( 4 \) means \( 2c+d\equiv0\pmod4 \); divisible by \( 8 \) means \( 4b+2c+d\equiv0\pmod8 \).
2. The pairs \( (c,d) \) with \( 2c+d\equiv0\pmod8 \) number \( 9 \): then \( 4b+0\not\equiv0 \) needs \( b \) odd (\( 5 \) choices). The pairs with \( 2c+d\equiv4\pmod8 \) number \( 9 \): then \( b \) must be even (\( 4 \) choices).
3. Digit \( a \) is free (\( 9 \) choices): total \( =9\times(9\times5+9\times4)=9\times81=729 \).

Teacher's Note:
1. The count follows from the last three digits: for each admissible last-two-digit pair there are \( 9 \) choices of the first digit and the hundreds digit is restricted by the \( 8 \)-condition.

Answer: (c) \( 144 \)

Explanation:
1. Positive solutions in total: \( \binom{19}{2}=171 \).
2. Solutions with at least two equal: \( x=y \) gives \( x=1,\dots,9 \) with \( z=20-2x \) (all \( 9 \) values positive, and \( z \) equal to \( x \) only when \( x=\frac{20}3 \) not an integer); by symmetry the three pair-equalities give \( 3\times9=27 \) solutions, with no overlap.
3. Unequal solutions: \( 171-27=144 \).

Teacher's Note:
1. Subtract the solutions in which two of the numbers are equal. Triple equality is impossible because \( 3\nmid20 \).

Answer: (b) \( 126 \)

Explanation:
1. \( S \) has \( 5 \) odd and \( 4 \) even numbers. \( N_k=\binom5k\binom4{5-k} \).
2. \( N_1+\dots+N_5 \) counts all \( 5 \)-element subsets that contain at least one odd number. Every \( 5 \)-subset has at least one odd (only \( 4 \) evens exist).
3. So the sum is \( \binom95=126 \).

Teacher's Note:
1. Since there are only four even numbers, every five-element subset has at least one odd element, so the sum is the total \( \binom95 \).

Answer: (d) \( 380 \)

Explanation:
1. No boy: \( \binom64=15 \) teams. Exactly one boy: choose \( 3 \) girls and \( 1 \) boy, \( \binom63\times4=80 \) teams.
2. Total teams \( =15+80=95 \); each can choose a captain in \( 4 \) ways: \( 95\times4=380 \).

Teacher's Note:
1. Count the teams first and then multiply by the number of choices of captain.