The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
A
$$7!\, \times 5!\,\,$$
B
$$6!\, \times 5!$$
C
$$30!$$
D
$$5!\, \times 4!$$
Explanation
6 men can sit at the round table = $$\left( {6 - 1} \right)! = 5!$$ ways
Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.
So total no of ways 6 men and 5 women can dine at the round table
= $$5!\, \times {}^6{C_5} \times 5!$$
= $$5!\, \times 6 \times 5!$$
= $$5!\, \times 6!$$
2
AIEEE 2002
MCQ (Single Correct Answer)
Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are
A
312
B
3125
C
120
D
216
Explanation
Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.
Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.
Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)
So total no of arrangement = 1$$ \times $$2$$ \times $$3$$ \times $$4$$ \times $$5 = 5!
Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)
So total no of arrangement = 4$$ \times $$4$$ \times $$3$$ \times $$2$$ \times $$1 = 4.4!
$$\therefore$$ Total arrangement = 5! + 4.4! = 216
3
AIEEE 2002
MCQ (Single Correct Answer)
The sum of integers from 1 to 100 that are divisible by 2 or 5 is
A
3000
B
3050
C
3600
D
3250
Explanation
According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$$ \times $$5 = 10.
Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100
This is an A.P where first term = 2, last term = 100 and total terms = 50.
$$ \therefore $$ Sum of the numbers divisible by 2
= $${{50} \over 2}\left[ {2 + 100} \right]$$
= 25$$ \times $$102
= 2550
Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100
$$ \therefore $$ Sum of the numbers divisible by 5
= $${{20} \over 2}\left[ {5 + 100} \right]$$
= 10$$ \times $$105
= 1050
And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100
$$ \therefore $$ Sum of the numbers divisible by 10
Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is
A
125
B
105
C
374
D
625
Explanation
There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed
Case 1 : First digit is 1 = 1 _ _ _
Possible numbers starting with 1 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125
But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.
Case 2 : First digit is 2 = 2 _ _ _
Possible numbers starting with 2 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125
Case 3 : First digit is 3 = 3 _ _ _
Possible numbers starting with 3 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125
Total possible numbers = 124 + 125 + 125 = 374
Questions Asked from Permutations and Combinations
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions