Test: Permutations And Combinations - 2 - Commerce MCQ

Sum of the numbers (S) formed by taking all the given n digits is-

S=(sum of all digits)×(n−1)!×(111....n times)

Now according to the question, the digits are 1,3,5,7

∴n = 4

Sum of digits = 1+3+5+7 = 16

∴S = 16×3!×1111

Hence the correct answer is  16×3!×1111.

′+′ signs can be put in a row in one way creating seven gaps shown as arrows:
Now 4′−′ signs must be kept in these gaps. So, no tow ′−′ signs should be together.
Out of these 7 gaps 4 can be chosen in 7C4 ways.

Number of tickets selected from first station =20
from second =19
.... for last station =0
We have to select 2 consecutive stations
so total number of possible tickets = P(20,2)

Take 1 person from 6 and fix him and 5 others can arranged in -- 5! ways=120

there are 6 places left in which 2 brothers can sit

so they can choose any 2 places from 6 - 6C2 ways=15

2 brothers can arrange themselves in 2! ways=15*2=30

total ways=120*30=3600

It is asked to find out the number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated.

According to the question, there are 8 flowers, but it is given that four particular flowers are never separated. So four particular flowers will always remain adjacent to each other.

To solve this problem we should know the following cases of circular permutations:

• If clockwise and anti-clockwise orders are different, the total number of circular permutations is: (n-1)!.
• If anti-clockwise and clockwise orders are taken as the same, then the total number of circular permutations is given by (n-1)!/2!

It is important to know which formula is applicable in this case.

When we turn a garland it forms a different arrangement. But it still is the same garland.
For example, consider a set of numbers like ‘1234’. When turned around it looks like ‘4321’ but it still is the same permutation of numbers.
In a similar way, the garland when turned might have a different permutation but it still is the same garland.
Therefore we’ll use the 2nd formula.
For n distinct objects in which clockwise and anticlockwise orders are taken as the same, the total number of circular permutations is: (n-1)!/2
Here in this case, given that four flowers must be together in all cases.
So we’ll club together all four particular flowers and consider this as one object.
Now we have 5 different objects - four flowers and four particular flowers clubbed as one object. Now we have to arrange these five objects in five circular seats.
In that case, the number of permutations is (5-1)!
And in every permutation, the clubbed flowers can be arranged in 4! ways.
So there are a total of (5-1)! * 4! Permutations.
But we have seen that the garland is the same in anticlockwise and clockwise order.
Therefore for every permutation, we have included its rotated permutation.
So the total number of circular permutations of garlands is : (5-1)!*4! /2

(5-1)! * 4! /2 = (4*3*2*1)*(4*3*2*1)/2
= 24 * 12 = 288

Thus the total number of ways in which 8 different flowers can be seated to form a garland so that four particular flowers are never separated is 288.

The mint has to perform two jobs:
1) Selecting the number of days in the February month i.e., 28 or 29
2) Selecting the first day of February month.
The first job can be completed in 2 ways while second job can be completed in 7 ways by selecting any one of the seven days of a week.
Thus, the required number of plates = 2 x 7 = 14.

 Number 1 is odd. As any number is divisible by 1: 1
Do prime factorisation of 3600: 2*2*2*2*3*3*5*5
Select all odd numbers from above: 3,3,5,5
Try every possible products of these:
Single number: 3, 5
Two numbers: 3*3, 3*5, 5*5: 9,15,25
Three numbers: 3*3*5, 3*5*5: 45, 75
All four: 3*3*5*5: 225
The odd divisors are: 1,3,5,9,15,25,45,75,225
There are 9 odd divisors of 3600.

All possible three number multiplications originate from the following triads:
1,1,30
1,2,15
1,3,10
1,5,6
2,3,5
First one can have 3!/2! = 3 ways and the remaining four triads can have 3! combinations
total combinations = 3 + 4*3! = 27

No.of choices for each question = 3
 
There are six questions so total = (3)¹²
So no.of ways = (3)¹² - 1

These digit number without digit 5 →100....999
→ these are 900 three-digit number
→ from 100 to 199 → 19 number with 5.
200−299→19
300−399→19
400−499→19
600−699→19
700−799→19
800−899→19
900−999→19
500−599→100
total number with 5=19×8+100 for (500-599)
 =152+100
 =252

First we fix the alternate position of the girls. Five girls can be seated around the circle in (5−1)!=4! , 5 boys can be seated in five -vacant place by 5!
∴ Required number of ways =4!×5!

2.6.10.14... to n factors = (2n)!/n! 
= 2(1.3.5.7………………25)
2.6.10.14... to 50 factors = (2*2*25)!/25! 
 = 100!/25!