JEE Mathematics Permutation and Combination MCQs Set 09

Answer: (a) \( \frac{7!}{2!}=2520 \)

Explanation:
1. Fix E at both ends. The remaining 7 letters are E, N, N, D, A, O, L (N repeated twice): \( \frac{7!}{2!}=2520 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 1001

Explanation:
1. Let each person get \( 2y_i+1 \) things: \( \sum(2y_i+1)=25 \) gives \( y_1+\dots+y_5=10 \).
2. Number of non-negative solutions = \( \binom{14}{4}=1001 \).

Teacher's Note:
1. Writing odd numbers as 2y + 1 makes the problem a standard stars-and-bars count.
2. Solutions of y1 + ... + y5 = 10 number C(14, 4).

Answer: (d) \( C_{10}-B_{10} \)

Explanation:
1. By Vandermonde, \( \sum_{r=0}^{10}A_rB_r=\sum\binom{10}{10-r}\binom{20}{r}=\binom{30}{10}=C_{10} \) and \( \sum_{r=0}^{10}A_r^2=\binom{20}{10}=B_{10} \).
2. Dropping the \( r=0 \) terms (each 1): \( B_{10}(C_{10}-1)-C_{10}(B_{10}-1)=C_{10}-B_{10} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 7

Explanation:
1. For \( n \) things: \( {}^nP_3-{}^nC_3=5\cdot{}^nC_3 \).
2. Need \( 5\,{}^nC_3>100 \), i.e. \( {}^nC_3>20 \).
3. \( {}^6C_3=20 \) (gives exactly 100, not more) and \( {}^7C_3=35 \), so the least n is 7.

Teacher's Note:
1. nPr = nCr x r!, so the difference is (3! - 1) times nCr.
2. Remember the difference must exceed, not equal, 100.

Answer: (d) 60

Explanation:
1. Even places: 2nd, 4th, 6th, 8th (4 places) for the odd digits 3, 3, 5, 5: \( \frac{4!}{2!\,2!}=6 \) ways.
2. The remaining 5 places hold 2, 2, 8, 8, 8: \( \frac{5!}{2!\,3!}=10 \) ways.
3. Total \( 6\times10=60 \).

Teacher's Note:
1. Place the odd digits first, then the others.
2. Divide by factorials for repeated digits.

Answer: (d) 4

Explanation:
1. Let the word have \( n \) letters: \( \frac{n!}{4!}=210 \), so \( n!=5040 \) and \( n=7 \).
2. The 4 identical letters count as one kind and the other 3 letters are distinct, so there are \( 3+1=4 \) different letters.

Teacher's Note:
1. The repeated letter is itself one of the different letters.
2. Solve n! = 5040 to get n = 7.

Answer: (d) \( 2^{4n}+{}^{4n+1}C_n \)

Explanation:
1. The terms \( {}^{4n+1}C_i \) for \( i=0 \) to \( n \) and the terms \( {}^{4n+1}C_j \) for \( j=n \) to \( 2n \) together cover \( j=0 \) to \( 2n \) once, with \( j=n \) counted twice.
2. Since \( {}^{4n+1}C_j={}^{4n+1}C_{4n+1-j} \), the sum of the first \( 2n+1 \) coefficients is half of \( 2^{4n+1} \), that is \( 2^{4n} \).
3. Total = \( 2^{4n}+{}^{4n+1}C_n \).

Teacher's Note:
1. Binomial coefficients are symmetric, so the first half of the row sums to half of the full row sum.
2. Check with \( n=1 \): the sum is 11 + 10 = 21, and \( 2^4+{}^5C_1=21 \).

Answer: (d) \( 7^{164} \)

Explanation:
1. Exponent \( =\left[\frac{1000}7\right]+\left[\frac{1000}{49}\right]+\left[\frac{1000}{343}\right]=142+20+2=164 \).

Teacher's Note:
1. Apply Legendre's formula.
2. Stop when the power of 7 exceeds 1000.

Answer: (c) 5! = 120

Explanation:
1. Treat ENDEA as one block. The remaining letters are N, O, E, L, so there are 5 objects (block, N, O, E, L), all distinct: \( 5!=120 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 41

Explanation:
1. Each element goes to the first subset, the second, or neither: \( 3^4=81 \) ordered pairs.
2. The only ordered pair with equal members is \( (\varnothing,\varnothing) \), so unordered pairs \( =\frac{81-1}{2}+1=41 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) (iii) and (iv) only

Explanation:
1. Only the terms \( 1,x,x^2 \) matter for low powers: \( a_1=10 \) (choose one factor to give x).
2. For \( x^2 \): either one factor gives \( x^2 \) (10 ways) or two factors give \( x \) each (\( \binom{10}{2}=45 \) ways), so \( a_2=55 \).
3. Hence \( a_1+a_2=65 \): (iii) and (iv) are correct.

Teacher's Note:
1. Count the ways of forming x and x squared from ten factors.
2. The same result follows from the expansion of (1 - x^5)^10 / (1 - x)^10.

Answer: (b) 7

Explanation:
1. \( T_n=\binom n3 \), so \( T_{n+1}-T_n=\binom n2=21 \) (Pascal's rule).
2. Then \( n(n-1)=42 \) and \( n=7 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac15(n-4) \)

Explanation:
1. \( T_5=\binom n4a^{n-4}b^4 \) and \( T_6=-\binom n5a^{n-5}b^5 \). Their sum is zero, so \( \binom n4a=\binom n5b \).
2. Hence \( \frac ab=\frac{\binom n5}{\binom n4}=\frac{n-4}{5} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 14

Explanation:
1. Total functions: \( 2^4=16 \). Non-onto ones are the 2 constant functions.
2. Onto functions: \( 16-2=14 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 96

Explanation:
1. Alphabetical order of the letters: C, C, H, I, N, O. Words start with C. Second letter C: the remaining H, I, N, O give \( 4!=24 \) words.
2. Second letter H, I or N (each smaller than O): \( 3\times4!=72 \) words.
3. Second letter O, third C, then H, I, N in order gives COCHIN itself. Words before it: \( 24+72=96 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 2 x 5! = 240

Explanation:
1. The odd positions (5 of them) hold A, E, E, E, O: \( \frac{5!}{3!}=20 \) ways.
2. The even positions (4) hold D, L, N, N: \( \frac{4!}{2!}=12 \) ways. Total \( 240=2\cdot5! \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 40

Explanation:
1. All arrangements: \( \frac{6!}{3!\,2!}=60 \).
2. Arrangements with the two Ns together (treat NN as one letter, letters B, A, A, A, NN): \( \frac{5!}{3!}=20 \).
3. Required \( 60-20=40 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 1

Explanation:
1. Let \( Q=\frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+\dots+b^{n-1} \) (n odd). The given sum is \( E=a^{n-1}+b^{n-1}-Q \).
2. By Fermat, \( a^n\equiv a,\ b^n\equiv b \) so \( (a+b)Q\equiv a+b \); as \( \gcd(a+b,n)=1 \), \( Q\equiv1 \).
3. Also \( a^{n-1}\equiv b^{n-1}\equiv1 \). So \( E\equiv1+1-1=1 \), i.e. a multiple of n plus 1.

Teacher's Note:
1. Express the sum through (a^n + b^n)/(a + b).
2. Fermat's little theorem gives a^(n-1) = 1 modulo n.

Answer: (d) 2 x 5! = 240

Explanation:
1. The last five positions can only hold letters from E, E, E, A, O: \( \frac{5!}{3!}=20 \) ways.
2. The first four positions hold D, L, N, N: \( \frac{4!}{2!}=12 \) ways. Total \( 20\times12=240=2\cdot5! \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \binom{n+2}{r} \)

Explanation:
1. Group: \( \left[\binom nr+\binom n{r-1}\right]+\left[\binom n{r-1}+\binom n{r-2}\right] \).
2. By Pascal's rule this is \( \binom{n+1}{r}+\binom{n+1}{r-1}=\binom{n+2}{r} \).

Teacher's Note:
1. Apply Pascal's rule twice.
2. Check with n = 2, r = 2: 1 + 4 + 1 = 6 = C(4,2).

Answer: (c) \( \frac{n(n+1)(n+2)}{6} \)

Explanation:
1. \( \frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}{r} \), so the term is \( r(n+1-r) \).
2. Sum \( =(n+1)\frac{n(n+1)}2-\frac{n(n+1)(2n+1)}6=\frac{n(n+1)(n+2)}6 \).

Teacher's Note:
1. Use the ratio of consecutive binomial coefficients.
2. Then use the standard sums of r and r squared.

Answer: (b) 24

Explanation:
1. Since \( y \) is even, \( x=50-y \) must be even.
2. Need \( y\ge2 \), so \( x\le48 \): \( x=2,4,\dots,48 \), giving 24 pairs.

Teacher's Note:
1. x must be in A (1 to 50) and y in B (even, from 2).
2. Count the even values of x from 2 to 48.

Answer: (b) 225

Explanation:
1. For each prime, the larger exponent among \( p,q \) must be the exponent in the LCM. For an exponent \( e \) the number of ordered pairs of exponents with maximum \( e \) is \( 2e+1 \).
2. So the count is \( (2\cdot2+1)(2\cdot4+1)(2\cdot2+1)=5\cdot9\cdot5=225 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 71

Explanation:
1. Each ball scores at most 6, so 30 runs in 6 balls means a deficit of 6 from the maximum 36.
2. Write the deficits per ball (\( 6-\text{runs} \)): possible deficits are 6, 5, 4, 3, 2, 0, and they must add to 6.
3. Count ordered 6-tuples: one deficit of 6 (6 ways); deficits 2 and 4 (6 x 5 = 30 ways); deficits 3 and 3 (15 ways); deficits 2, 2, 2 (20 ways).
4. Total = 6 + 30 + 15 + 20 = 71.

Teacher's Note:
1. Think of how far each ball falls short of 6.
2. Group the cases by which deficits are used and count positions.

Answer: (c) (i) and (iii) only

Explanation:
1. \( 17\equiv1\pmod{16} \), so \( 17^{22}\equiv1 \): divisible by 16.
2. By Fermat, \( 17^{22}\equiv1\pmod{23} \) and the number is even, so it is divisible by 46.
3. Mod 11: \( 17^{22}\equiv6^{22}\equiv6^2=36\equiv3 \), so not divisible by 44. Mod 27: \( 17^{22}\bmod27=10 \) (not 0), so 27 does not divide it.

Teacher's Note:
1. Check each divisor with congruences.
2. Fermat's theorem: a^(p-1) is 1 modulo p for p prime not dividing a.

Answer: (c) 77

Explanation:
1. Let the digits 1, 2, 3 occur \( a,b,c \) times: \( a+b+c=7,\ a+2b+3c=10 \) so \( b+2c=3 \).
2. Cases: \( (a,b,c)=(4,3,0) \) gives \( \frac{7!}{4!3!}=35 \); \( (5,1,1) \) gives \( \frac{7!}{5!}=42 \).
3. Total \( 35+42=77 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 15

Explanation:
1. By Vandermonde's identity the sum equals \( \binom{30}{m} \).
2. This is largest at the middle value \( m=15 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.