JEE Mathematics Permutation and Combination MCQs Set 13

Answer: (a) \( {}^{13}C_6 \)

Explanation:
1. We need non-negative solutions of \( x_1+\dots+x_8=6 \).
2. Number of solutions \( \binom{6+7}{7}=\binom{13}{6} \).

Teacher's Note:
1. Selecting with repetition from n kinds, r items: C(n + r - 1, r).
2. Here n = 8 and r = 6.

Answer: (d) 4

Explanation:
1. Odd letters L1, L3, L5 go into E1, E3, E5 as a derangement of 3 objects: \( D_3=2 \).
2. Even letters likewise: \( D_3=2 \).
3. Total \( 2\times2=4 \).

Teacher's Note:
1. The two groups are independent.
2. The derangement numbers are D2 = 1, D3 = 2, D4 = 9.

Answer: (a) 53

Explanation:
1. Cards 3, 4, 5, 6, 7 remain for envelopes 1, 4, 5, 6, 7. Cards 4 to 7 must avoid their own envelopes, while card 3 is free (its own envelope 3 is already taken).
2. If card 3 goes to envelope 1, the other four cards must be deranged: \( D_4=9 \) ways.
3. If card 3 does not go to envelope 1, treat envelope 1 as the 'own' envelope of card 3 and derange all five: \( D_5=44 \) ways.
4. Total \( 44+9=53 \).

Teacher's Note:
1. Think of envelope 1 as the 'home' of card 3, then count cases.
2. D_5 = 44 and D_4 = 9.

Answer: (d) \( \frac{(m+n-2)!}{(m-1)!\,(n-1)!} \)

Explanation:
1. The grid has \( m-1 \) blocks in one direction and \( n-1 \) in the other.
2. A shortest route has \( m+n-2 \) steps, of which \( m-1 \) are in one direction.
3. Number of routes \( \frac{(m+n-2)!}{(m-1)!\,(n-1)!} \).

Teacher's Note:
1. Count blocks, not streets.
2. Arrange the steps of two kinds in a row.

Answer: (c) 49

Explanation:
1. Two squares share exactly one corner when they are diagonal neighbours; such squares have the same colour.
2. Number of diagonal pairs on the board: along each of the two diagonal directions there are \( 7\times7=49 \) pairs, so 98 pairs in all, and these split equally between black and white squares.
3. Black pairs \( 98/2=49 \).

Teacher's Note:
1. Diagonally adjacent squares are of the same colour.
2. Count pairs of squares in a 2 x 2 block along each diagonal.

Answer: (c) 1981

Explanation:
1. All non-empty selections: \( 2^{11}-1=2047 \).
2. Remove selections of 1 or 2 flowers: \( 11+\binom{11}{2}=11+55=66 \).
3. Required \( 2047-66=1981 \).

Teacher's Note:
1. Total subsets minus the unwanted small subsets.
2. Include the empty set in the removal.

Answer: (b) \( 45\cdot(55)^2 \)

Explanation:
1. No borrowing means each digit of \( \alpha \) is at most the matching digit of \( \beta \).
2. Hundreds digits: \( 1\le x_1\le y_1\le9 \) gives \( 45 \) pairs. Tens digits: \( 0\le x_2\le y_2\le9 \) gives \( 55 \) pairs; likewise units: 55.
3. Total \( 45\times55\times55=45\cdot(55)^2 \).

Teacher's Note:
1. Count pairs digit by digit.
2. The hundreds digit cannot be zero.

Answer: (c) \( \frac18n(n-1)(n-2)(n-3) \)

Explanation:
1. The number of intersection points is \( \binom n2 \), so lines joining pairs of them number \( \binom{\binom n2}{2} \).
2. Each original line contains \( n-1 \) intersection points and so accounts for \( \binom{n-1}{2} \) of those joins, which are not fresh.
3. Fresh lines \( \binom{\binom n2}{2}-n\binom{n-1}{2}=\frac18n(n-1)(n-2)(n-3) \).

Teacher's Note:
1. Subtract the joins that lie on the original lines.
2. Test with n = 4: the answer is 3.

Answer: (a) \( \frac{(4^{2017}-3^{2017})+(2^{2017}-1)}{2} \)

Explanation:
1. Ordered pairs with \( A\cup B\ne X \): all \( 4^{2017} \) minus the \( 3^{2017} \) pairs with union \( X \).
2. Pairs with \( A=B \) and \( A\ne X \): \( 2^{2017}-1 \).
3. Unordered pairs \( \frac{(4^{2017}-3^{2017})+(2^{2017}-1)}{2} \).

Teacher's Note:
1. Convert ordered to unordered by adding the symmetric cases and halving.
2. Each element has 4 choices: in A only, B only, both, neither.

Answer: (d) \( 12!\times2 \)

Explanation:
1. Same paper in a column and different papers side by side force the pattern ABABAB in both rows, or BABABA.
2. So there are 2 patterns of papers over the 12 seats.
3. For each pattern the 12 different students can be seated in \( 12! \) ways. Total \( 12!\times2 \).

Teacher's Note:
1. Find the paper pattern first, then place the students.
2. Students are distinct even if papers are alike.

Answer: (a) \( \frac{6\times6!}{1!2!3!}+3\times\frac{6!}{3!3!}+\frac{6!}{2!2!2!} \)

Explanation:
1. The numbers of invitations (in some order) are 3-2-1, 3-3-0 or 2-2-2.
2. 3-2-1: the counts can be assigned to the friends in \( 3!=6 \) ways, each giving \( \frac{6!}{3!2!1!} \) sequences. 3-3-0: choose the friend not invited in 3 ways, \( \frac{6!}{3!3!} \) each. 2-2-2: \( \frac{6!}{2!2!2!} \).
3. Total \( 6\times60+3\times20+90=510 \), which is option (a).

Teacher's Note:
1. List the possible partitions of 6 into three parts each at most 3.
2. Multiply by the number of ways to assign the counts to friends.

Answer: (a) \( \frac{36!\cdot4!}{(9!)^4} \)

Explanation:
1. Choose which suit's honours go to which player: \( 4! \) ways.
2. The remaining 36 cards are shared equally, 9 to each player: \( \frac{36!}{(9!)^4} \).
3. Total \( \frac{36!\cdot4!}{(9!)^4} \).

Teacher's Note:
1. Distribute the special cards first.
2. Each player still needs 9 more cards.

Answer: (a) 685

Explanation:
1. For sum \( s \) the number of positive solutions is \( \binom{s-1}{2} \).
2. Required \( \binom{15}{2}+\binom{16}{2}+\binom{17}{2}+\binom{18}{2}+\binom{19}{2}=105+120+136+153+171=685 \).

Teacher's Note:
1. Add the cases s = 16, 17, 18, 19, 20.
2. Positive solutions of a sum s: C(s - 1, 2).

Answer: (a) \( {}^6C_3-7 \)

Explanation:
1. Total triples \( \binom63=20 \).
2. Triples failing the triangle inequality: (2,3,5), (2,3,6), (2,3,7), (2,4,6), (2,4,7), (2,5,7), (3,4,7), i.e. 7.
3. Triangles \( 20-7=13 \).

Teacher's Note:
1. A triple forms a triangle if the two smaller lengths add to more than the largest.
2. List the failing triples systematically.

Answer: (a) \( {}^nP_2\cdot{}^mP_m\cdot(n-2)! \)

Explanation:
1. \( P={}^nP_2\cdot(m+n-2)! \) and \( Q=\frac{(m+n-2)!}{m!\,(n-2)!} \).
2. So \( \frac PQ={}^nP_2\cdot m!\,(n-2)!={}^nP_2\cdot{}^mP_m\cdot(n-2)! \).

Teacher's Note:
1. Choose the two end oranges, then arrange the rest.
2. For alike fruits divide by the factorial of each species count.

Answer: (a) 256

Explanation:
1. \( 25! \) has 9 distinct prime factors (2, 3, 5, 7, 11, 13, 17, 19, 23).
2. Each prime's full power goes to \( m \) or to \( n \): \( 2^9=512 \) ordered pairs.
3. Exactly half have \( m\lt n \): \( 256 \).

Teacher's Note:
1. Coprime factorisation: each prime goes wholly to one side.
2. The fraction must be less than 1, so halve the ordered pairs.

Answer: (a) 1.17

Explanation:
1. Divisible by 3: digit sum must be a multiple of 3; counting gives \( M=744 \).
2. Divisible by 4: the last two digits must form a multiple of 4; counting gives \( N=636 \).
3. So \( \frac MN=\frac{744}{636}\approx1.17 \).

Teacher's Note:
1. Count by cases on the last digits for 4 and on digit sets for 3.
2. This count is easy to confirm by listing sets of digits with sum divisible by 3.

Answer: (c) 8

Explanation:
1. The numbers of coins must be 1, 2, 4 in some order: \( 3!=6 \) assignments.
2. Each gives \( \frac{7!}{1!2!4!}=105 \) divisions, so \( k=630=2\cdot3^2\cdot5\cdot7 \).
3. With 4 distinct primes the coprime splittings number \( 2^{4-1}=8 \).

Teacher's Note:
1. The only partition of 7 into three distinct positive parts is 1 + 2 + 4.
2. Count coprime splits via the distinct prime factors.

Answer: (c) 10

Explanation:
1. The vowels can be arranged in \( 5! \) ways and the digits in \( 10! \) ways.
2. The four gaps hold odd numbers of digits \( 2a_i+1 \) with total 10, so \( a_1+a_2+a_3+a_4=3 \): \( \binom63=20 \) ways.
3. Total \( 20\cdot10!\cdot5! \), so \( b=10 \).

Teacher's Note:
1. Write each odd gap as 2a + 1 to convert to a stars-and-bars count.
2. Arrange vowels and digits separately.

Answer: (b) 46.20

Explanation:
1. The possible shares are 4-3-1, 4-2-2 and 3-3-2 in some order.
2. 4-3-1: \( 3!\times\frac{8!}{4!3!1!}=6\times280=1680 \). 4-2-2: \( 3\times\frac{8!}{4!2!2!}=3\times420=1260 \). 3-3-2: \( 3\times\frac{8!}{3!3!2!}=3\times560=1680 \).
3. Total \( N=4620 \), so \( N/100=46.20 \).

Teacher's Note:
1. List the allowed partitions of 8 into three parts between 1 and 4.
2. Count the ways to give the shares to specific boys.

Answer: (b) 1.90

Explanation:
1. If Y wins \( k \) matches (0 to 9), the last match is won by X: \( N=\sum_{k=0}^{9}\binom{9+k}{k}=\binom{19}{9}=92378 \).
2. With X winning the first match: \( M=\sum_{k=0}^{9}\binom{8+k}{k}=\binom{18}{9}=48620 \).
3. Ratio \( \frac{92378}{48620}=1.90 \).

Teacher's Note:
1. Use the hockey-stick identity for the sums.
2. The last match of the series is always won by the series winner.

Answer: (d) 30

Explanation:
1. Let the numbers of 4-pointers, 2-pointers and 0's be \( a,b,c \) with \( 4a+2b=14 \) and \( a+b+c=5 \).
2. Possibilities: \( (a,b,c)=(3,1,1) \) gives \( \frac{5!}{3!1!1!}=20 \); \( (2,3,0) \) gives \( \frac{5!}{2!3!}=10 \).
3. Total \( 20+10=30 \).

Teacher's Note:
1. List the possible numbers of each score first.
2. Arrange the shots with repeated scores.