JEE Main Mathematics Permutations and Combinations Online Test

Number of ways to fill the IIIrd place = 8

Number of ways to fill the IVth place = 7

Number of ways to fill the Vth place = 6

Hence, by FPC total number of ways = 8 × 7 × 6 = 56 × 6 = 336

We cannot write 903

If in the selection of n cards, we get either

(9 or 3), (9 or 0), (0 or 3), (only 0), (only 3) or (only 9)

Can be selected = 2 × 2 × 2 × ……… n times = 2ⁿ

Similarly, (9 or 0) or (3 or 0) can be selected 2ⁿ

In the above selection (only 0) or (only 3) or (only 9) is repeated twice

∴ Total ways = 2ⁿ + 2ⁿ + 2ⁿ – 3 = 93

⇒ 3.2ⁿ = 96

⇒ 2ⁿ = 32 = 2⁵

∴         n = 5

⁴C₂ × ⁵C₃ × 5! = 7200

2⁵ × 2⁴ × 2³ – 1 = 2¹² – 1 = 4095

In a word TRIANGLE, vowels are (A,E, I) and consonants are (G, L, N, R, T).

First, we fix the 5 consonants in alternate position in 5! ways.

_G_L_N_R_T_

In rest of the six blank positions, three vowels can be arranged in ⁶P₃ ways.

∴ Total number of different words = 5! × ⁶P₃

= 120 × 6 × 5 × 4 = 14400

The number of words starting from A are 5! = 120

The number of words starting from 1 are = 5! = 120

The number of words starting from KA are = 4! = 24

The number of words starting from KI are = 4! = 24

The number of words starting from KN are 4! = 24

The number of words starting from KRA are = 3! = 6

The number of words starting from KRIA are 2! = 2

The number of words starting from KRIN are  = 2! = 2

The number of words starting from KRISA are = 1! = 1

The number of words starting from KRISNA are = 1! = 1

Hence, rank of the word KRISNA

= 2(120) + 3 (24) + 6 + 2(24) + 6 + 2(2) + (1) = 324

9600 = 2⁷ × 3 × 5²

∴ No. of divisors = (7 + 1) × (1 + 1) × (2 + 1) = 8 × 2 × 3 = 48

2¹⁰ – 1 = 1023

Case I when number of two digits.

Total number of ways = ⁹C₁ × ⁹C₁ = 9 × 9  = 81

Case II when number in three digits

Total number of ways = ⁹C₁ × ⁹C₁ × ⁹C₁ = 9 × 9 × 9 = 729

∴ Total number of ways = 81 + 729 = 810

⁸C₂ –8 = 28 – 8 = 20

∴ Required number of ways = ^22–4–2C_11–2 = ¹⁶C₉

We can think of three packets. One consisting of three boys of class X, other consisting of four boys of class XI and last one consisting of five boys of class XII. These packets can be arranged in 3! ways and contents of these packets can be further arranged in 3!4! and 5! ways, respectively. Hence, the total number of ways is 3! × 3! × 4! × 5!

Suppose a₁a₂a₃a₄a₅a₆a₇ represents a seven digit number. Then a₁ takes the value 1, 2, 3, …… , 9 and a₂,a₃, ….. , a₇ all take values 0, 1, 2, 3, …., 9. If we keep a₁a₂a₃ …. a₆ fixed, then the sum a₁ +a₂ + a₃ + … …. + a₆ is either even or odd. Since a₇ takes 10 values 0, 1, 2, ……, 9 five of the numbers so formed will be even and five odd.

Hence, the required number of numbers

= 9∙10 ∙10 ∙10∙10∙10∙5 = 45 × 10⁵

The sum of the digit in unit place of all the numbers formed = 3! (3 + 4 + 5 + 6) = 6 × 18

= 108

Each place can be filled in 3 ways

∴ n places can be filled in 3ⁿ

∴ 3ⁿ ≥ 900

∴ 3⁶ = 729 and 3⁷ = 2187

Hence              n = 7

x, 1, 2, are boys and y, 3, 4 are girls

No. of arrangement = 2! × 2! = 4

The number of ways in which we can choose a committee

= Choose two men and four women + choose three men and six women

= ⁴C₂ ×⁶C₄ + ⁴C₃ × ⁶C₆

= 6 × 15 + 4 × 1 = 90 + 4 = 94

The number of ways that the candidate may select

(i) if 2 questions from A and 4 questions from B = ⁵C₂ × ⁵C₄ = 50

(ii) 3 questions from A and 3 questions from B = ⁵C₂ × ⁵C₃ = 100

and (iii) 4 questions from A and 2 questions from B = ⁵C₄ × ⁵C₂ = 50

Hence, total number of ways = 50 + 100 + 50 = 200

Coeff. of x⁴ in : (1 + x + x² + x³ + x⁴ + x⁵) (1 + x +x² + x³ +x⁴) (1 +x + x²)

: (1–x⁶) (1 –x⁵) (1–x³) (1–x)^–3

: [1–x³] [1 – x]^–3

: [^3+4–1C₄ – ^3+4–1C₁]

: 15 – 3

: 12

First stall can be filled in 3 ways, second stall can be filled in 3 ways and so on.

∴ Number of ways of loading steamer

= ³C₁ × ³C₁×  …..׳C₁ (12 times)

= 3 × 3 × ……× 3 (12 times) = 3¹²

Total ways = (8C2 –2) × ⌊6

= (28 – 2) × ⌊6

= 26 × ⌊6

Treat India and Pakistan prime ministers as one (I, P) + 7 others we have to arrange 8 person round a table no. of ways = 7! but corresponding to each arrangement India and Pakistan prime ministers total way = 7!.2!

If the function is one one, then select any three from the set B in ⁷C₃ ways i.e., 35 ways.

If the function is many one, then there are two possibilities, All three corresponds to same element number of such functions = ⁷C₁ = 7 ways. Two corresponds to same element. Select any two from the set B. The larger one corresponds to the larger and the smaller one corresponds to the smaller the third may corresponds to any two. Number of such functions = ⁷C₂ × 2 = 42

So, the required number of mapping = 354 + 7 + 42 = 84

Words beginning with A, D, M, N and O are 5! each. Words beginning with RAD are 3! and also for RAM are 3!. Then comes RAND. First we shall have RANDMO and the RANDOM.

∴ rank is 5(5!) + 2(3!) + 2 = 614

Number of lines from 6 point = ⁶C₂ = 15

Points of intersection obtained from these lines = ¹⁵C₂ = 105

Now, we find the number of times, the original 6 points come.

Consider one point say A₁. Joining A₁ to remaining 5 points, we get 5 lines and any two lines from these 5 lines gives A₁ as the point of intersection.

∴ A₁ is common in ⁵C₂ = 10 times out of 105 points of intersections.

Similar is the case with other five points.

∴ 6 original points come 6 × 10 = 60 times in points of intersection.

Hence, the number of distinct points of intersection

= 105 – 60 + 6 = 51

Case-I If B is right on A

Sub case –I C is right on B

then no. of ways = (4 – 2)! = 6

Sub case-II if D is right on B

then no. of ways = (4 – 1)! = 6

Case-II if C is right on A

⇒ D must be right on B = (4 – 1)! = 3! = 6

Hence total no. of ways is 6 + 6 + 6 = 18

Number of parallelograms = ⁵C₂ × ⁴C₂ = 60

Total number of points are m + n + k,  the triangles formed by these points = ^{m + n +k}C₃

Joining of three points on the same line gives no triangle, the number of such triangles is ^mC₃ + ⁿC₃ + ^kC₃

∴ Required number of triangles = ^{m + n + k}C₃ – ^mC₃ – ⁿC₃ – ^kC₃

Each question can be answered in 4 ways and all question can be answered correctly in only one way.

So, required number of ways = 63