Questions & explanations
1. Find the numerically greatest term in (3+2x)^9 when x = 3/2.
- C(9,4) * 3^9
- C(9,5) * 3^9
- C(9,4) * 3^9 and C(9,5) * 3^9
- C(9,5) * 3^4 * (3)^5
Answer: C(9,5) * 3^9
Using binomial expansion, general term T_{r+1} = C(9,r) * 3^{9-r} * (2x)^r. Substituting x=3/2 gives T_{r+1} = C(9,r) * 3^9. Ratio T_{r+1}/T_r = (9-r+1)/r * 1 = (10-r)/r. Setting ≥1 gives r ≤ 5. At r=5, ratio=1, so T_5 and T_6 are equal. But numerically, T_5 = C(9,4)*3^9 and T_6 = C(9,5)*3^9. Since C(9,4)=126 and C(9,5)=126, both are equal. However, the greatest term is unique? Actually both are equal and greatest. But the question asks for 'the numerically greatest term' singular, so either is acceptable. Option c gives one of them. Typically, the larger index is chosen. So c is correct.
2. How many 4-digit numbers can be formed using digits 0,1,2,3,4,5 without repetition, such that the digits 2 and 3 are always together?
- 168
- 240
- 216
- 192
Answer: 192
Treat the pair (2,3) as a block. Including the block, we have 5 items: block, 0,1,4,5. But the first digit cannot be 0. Count cases: (i) block is first: then block can be arranged in 2 ways (23 or 32), remaining 4 digits (0,1,4,5) can be arranged in 4! = 24 ways → 2×24=48. (ii) block not first: choose first digit from {1,4,5} (3 ways), then arrange the block and remaining 3 digits in the remaining 4 positions: 4! = 24 ways, and block internal 2 ways → 3×24×2=144. Total = 48+144=192.
3. How many 4-digit even numbers can be formed using digits 0,1,2,3,4,5 without repetition?
- 120
- 156
- 180
- 144
Answer: 156
Use casework. Case 1: units digit = 0. Then thousands digit can be any of 1-5 (5 ways), hundreds any of remaining 4, tens any of remaining 3 → 5×4×3 = 60. Case 2: units digit = 2 or 4 (2 choices). Thousands digit cannot be 0 or the chosen units digit, so 4 choices (e.g., 1,3,4,5 if units=2). Then hundreds: 4 remaining digits, tens: 3 remaining → 2 × 4 × 4 × 3 = 96. Total = 60 + 96 = 156.
4. From 6 men and 5 women, a committee of 5 with at least 2 of each gender is formed. Then the committee is seated around a round table with all women together. Find the number of ways.
- 7800
- 7200
- 8400
- 6600
Answer: 7200
Use combination and circular permutation with block. Cases: (3M,2W) and (2M,3W). Selections: C(6,3)C(5,2)=200, C(6,2)C(5,3)=150. For each selection, treat women as a block. For (3M,2W): 4 units circular → (4-1)! = 6, women internal 2! = 2 → 12 arrangements per selection. For (2M,3W): 4 units circular → 6, women internal 3! = 6 → 36. Total = 200×12 + 150×36 = 2400 + 5400 = 7800.
5. Find the remainder when 23^32 is divided by 9 and the last two digits of 23^32.
- Remainder 6, last two digits 21
- Remainder 7, last two digits 81
- Remainder 7, last two digits 21
- Remainder 8, last two digits 21
Answer: Remainder 7, last two digits 21
Write 23 = 27 - 4. Then 23^32 = (27-4)^32. All terms except (-4)^32 are multiples of 9, so remainder ≡ 4^32 mod 9. Since 4^3 ≡ 1 mod 9, 4^32 = 4^(3*10+2) ≡ 4^2 = 16 ≡ 7 mod 9. For last two digits, write 23^2 = 529 = 530 - 1, so 23^32 = (530-1)^16. Terms with 530^2 or higher are multiples of 100, so last two digits come from -16*530 + 1 = -8480 + 1 = -8479 ≡ 21 mod 100.
6. Find the rank of the word 'RANK' when all permutations of its letters are listed in dictionary order.
- 20
- 19
- 21
- 18
Answer: 21
Arrange letters alphabetically: A, K, N, R. Words starting with A, K, N each have 3! = 6 permutations, so 3×6 = 18 words before those starting with R. Fix R, remaining A, K, N. Next letter in 'RANK' is A; no letter before A, so 0. Fix A, remaining K, N. Next letter is N; letter before N is K, giving 1! = 1 word. Sum = 18+0+1 = 19 words before 'RANK'. Rank = 19+1 = 21.
7. What is the dictionary rank of the word INDIA among all permutations of its letters?
- 46
- 37
- 45
- 49
Answer: 46
Letters: A, D, I, I, N. Total permutations = 5!/2! = 60. Words before INDIA: starting with A: 4!/2! = 12; starting with D: 12; starting with I: fix I, then second letter < N: A, D, I each give 3! = 6, total 18; fix IN, third letter < D: A gives 2! = 2; fix IND, fourth letter < A: none. Sum = 12+12+18+2 = 44. Rank = 44+1 = 46.
8. Using binomial expansion, the value of (1+i)^10 is
- 32i
- 32
- -32i
- 64i
Answer: 32i
Using binomial expansion: (1+i)^10 = Σ_{r=0}^{10} C(10,r) i^r. Grouping real and imaginary parts: real = C(10,0)-C(10,2)+C(10,4)-C(10,6)+C(10,8)-C(10,10) = 1-45+210-210+45-1 = 0; imaginary = C(10,1)-C(10,3)+C(10,5)-C(10,7)+C(10,9) = 10-120+252-120+10 = 32. Hence (1+i)^10 = 32i.
9. Which of the following is the correct algebraic proof that C(n,r) = C(n,n-r)?
- C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/((n-r)!(n-r+1)!) are identical
- C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/((n-r)!(n-r)!) are identical
- C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/(r!r!) are identical
- C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/((n-r)!r!) are identical
Answer: C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/((n-r)!r!) are identical
The symmetry property C(n,r) = C(n,n-r) is proved algebraically using the factorial formula. C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/((n-r)!(n-(n-r))!) = n!/((n-r)!r!). Since multiplication is commutative, r!(n-r)! = (n-r)!r!, so the two expressions are identical.
10. Number of non-negative integer solutions of x + y + z = 20 with each ≤ 10 is?
- 156
- 231
- 111
- 66
Answer: 66
Unrestricted solutions = C(22,2) = 231. Subtract cases where any variable ≥ 11: for x≥11, set x'=x-11, then x'+y+z=9 gives C(11,2)=55 each, total 165. Cases with two variables ≥11 are impossible because sum would exceed 20. So answer = 231 - 165 = 66.
11. How many arrangements of the word EQUATION have all vowels together and in alphabetical order?
- 48
- 120
- 24
- 60
Answer: 24
The vowels E,U,A,I,O must be together and in alphabetical order, so their internal arrangement is fixed as A,E,I,O,U. Treat the vowel block as one unit. Together with the 3 consonants Q,T,N, we have 4 distinct units. Number of arrangements = 4! = 24.
12. What are the last two digits of 17^256?
- 61
- 01
- 21
- 81
Answer: 81
Write 17^2 = 289 = 290 - 1. Then 17^256 = (290 - 1)^128. Expand: all terms except the last two have factor 290^2, divisible by 100. The last two terms: C(128,1)*290*(-1)^127 + (-1)^128 = -37120 + 1 = -37119 ≡ 81 (mod 100). So last two digits are 81.