JEE Questions for Maths Permutations And Combinations Quiz 4 - MCQExams.com

3 men and 2 women equal to 5. A group of 5 members make 5! Permutation with each other. Therefore , the number of ways to sit 5 members = 5!. 6 places are filled  by 5 members by ⁶C₅ ways .Therefore, total number of ways to sit 5 members on 6 seats of a bus = ⁶C₅ × 5!.

Required number of ways = ⁴C₃ = 4

Required number of ways = ⁴C₂  × ³C₂ = 18.

Number of lines from 6 points = ⁶C₂ = 15. Points of intersection obtained from these lines  = ¹⁵C₂ = 105. Now we find the number of times, the original 6points come.

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

Therefore, A₁ come  ⁵C₂ = 10 times is 105 points of intersections. Similar is the case with other five points.

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

Hence the number of distinct points of intersection = 105 – 60 + 6 = 51.

Each set is having  m + 2 parallel lines and each parallelogram is formed by choosing two straight lines from the first set and two straight lines from the  second set. Two straight lines from the first set can be chosen in ^m+2C₂ ways and two straight lines from the second set can be chosen in ^m+2C₂ ways.

Hence the total number of parallelograms formed

= ^m+2C₂ . ^m+2C₂ = ( ^m+2C₂ )² .

Between 10 and 1000, the numbers are of 2 digits and 3 digits.

Since repetition is allowed , so each digit can be filled in 9 ways.

Therefore, number of 2 digit numbers = 9 × 9 = 81

and number of 3 digit numbers 9×9×9=729

Hence total numbers ways = 81 + 729 = 810

Number between 1 to 1000 will be one digit, two digit and three digit nos. Hence no. of single digit nos. will be =1.

Number of two digit no. will be

1. When 3 is at the unit place then tens place can be filled by 9 ways hence 9 ×      1  = 9 ways .
2. When 3 is at the tens place then unit place can be filled by 10 ways hence 1 × 10 = 10 ways .

Number of 3 digit nos.

1. 9  10  1 = 9 × 10 = 9 ways
2. 9  10  1 = 9 × 10 = 9 ways
3. 9  10  1 = 9 × 10 = 9 ways

So 100 + 90 + 90 = 280

Total nos. of times  in which 3 is used = 1 + 19 + 280 = 300.

Let E(n) denote the exponent of 3 in n. The greatest integer less than 100 divisible by 3 is 99.

We have E(100!) = E(1.2.3.4…..99.100)

= E(3.6.9…99) = E (3.1)(3.2)(3.3)…..(3.33)

= 33 + E(1.2.3……..33)

Now E (1.2.3……33) = E (3.6.9……33)

= E [(3.1) (3.2)(3.3)….(3.11)]

= 11 + E (1.2.3……11)

and  E (1.2.3……11) = E(3.6.9)

= E [(3.1)(3.2)(3.3)]

3 + E(1.2.3) = 3 + 1 = 4

Thus E (100!) = 33 + 11 + 4 = 48.

Each child will go as often as he (or she ) can be accompanied by two others.

Therefore, the required number is ⁷C₂ = 21.

Since 2 persons can drive the car, therefore, we have to select 1 from these two. This can be done in ²C₁ ways.Now from the remaining 5 persons we have to select 2 which can be done in ⁵C₂ ways. But the front seat and the rear seat person can interchange among themselves. Therefore, the required number of ways in which the car can be filled is

⁵C₂ × ²C₁ × 2! = 20 × 2 = 40.

12 persons can be seated around a round table in 11! ways. The total number of ways in which 2 particular persons sit side by side is 10! × 2!. Hence the required number of arrangements = 11! – 10! × 2! = 9 × (10!).

We have , 30 = 3 × 2 × 5.So, ‘a’ can take any value from 2,3,5 in 3 ways , ‘b’ can take the value in 2 ways and ‘c’ in 1 ways.

Therefore, Number of solution  = 3 × 2 × 1 = 6.

The arrangement can be done in ⁴P₄ = 24ways.

Total number of distinct functions from A to A are n^r i.e., 10¹⁰.

All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore, number of 5 digit numbers = ⁶P₅ – ⁵P₅ = 600.

[Since the case that 0 will be at ten thousand place should be omit]

Similarly number of 6 digit numbers

6! – 5! = 600

Now, the numbers of 4 digit numbers which  are greater than 3000, having  3,4 or 5 at first place, this can be done in 3 ways and remaining 3 digit may be filled from remaining 5 digit i.e., required number of 4 digit numbers are 5P3 ×3 = 180.Hence total required number of numbers = 600  +  600  + 180 = 1380.

2¹⁰  - 1=1023, -1 corresponds to none of the lamps is being switched on.

Suppose  x₁ x₂ x₃ x₄ x₅ x₆  x₇ represents a seven digit number. Then x₁ takes the value 1,2,3,….9 and  x₂,x₃,…..x₇  all take values 0,1,2,3,…..9. If we keep x₁,x₂,…x₆ fixed , then the sum x₁ + x₂ +….+ x₆ is either even or odd. Since  x₇ takes 10 values 0,1,2,…,9, five of the numbers so formed will be even and 5 odd.

Hence the required number of numbers

= 9.10.10.10.10.10.5 = 4500000.