Permutations Non Distinct Objects - Free MCQ Practice Test with solutions,

1 balloon is fixed that is to be placed above the cake
Now she have to arrange 9 balloons = 9!/(5!*2!*2!)
(9*8*7*6*5!)/((5!*2*2)
= (9*8*7*6)/(2*2)
= 756

The first letter can be posted in 4 ways. So, total outcomes about the first letter = 4.
For every outcome about the first letter, the second letter can be posted in 4 ways. So, total outcomes about the first and the second letters= (4*4) = 16.
Therefore, following the same route, we can say, total possible outcomes about the first and the second letters= (4*4*4) = 64.

(n + 1)! = 20 (n – 1)!
n (n + 1) = 20
(n – 4) (n + 5) = 0          
Since, (n – 1)! exists, n ≥ 1
So, n = 4 

Number of permutations of n different things taken r at a time when repetition is allowed = n^r

In a class of 6 students. Two students at both ends. Remaining four students have to be arranged 
= 2 * 4! 
= 2 * 24
= 48

If the first & last letter is fixed, then we find out, the number of permutations of the remaining letters, i.e. 4
= 4!/2!
= 4*3*2!/2!
= 12

Sweet has 5 letters, so it gives 5! but there are two letters which are same
Number of words will be = (5! ÷ 2!) = 60

n!/(n-r)!
4!/(4-2)!  = 4*3*2!/2!
= 12

3 white balls, 3 red balls
Total balls = 6
Total no. of ways = (6!)/(3!)(3!)
= (6*5*4*3!)/(3!*3*2)
= (5*4) = 20

1/4! + 1/5!
= 1/4![1 + 1/5]
1/4![6/5]
=> 6/5!

At least one yellow chair means “total way no yellow chair
Total ways to select 3 chairs form the total (2+3+4) chairs ⁹C₃
⁹C₃ = 9!/3!*6! = 84ways
Now, we dont want even one yellow chair
So, we should select 3 chairs from 6 chairs (2green & 4blue) = ⁶C₃
⁶C₃ = 6!/3!3! = 20 ways
hence , ways to select at least one yellow chair = 84-20 = 64 ways.

 If you imagine there are 10 spots on the floor in a line where people can stand, then the first person has 10 choices of where to stand.
The second person has 9 choices since the first person is already on one of the spots.
The third person has 8 choices. And so on until the last person has no choice.
We can now mulitply the numbers together. 10x9x8x7x6x5x4x3x2x1. This is called a factorial (where you multiply all of the consecutive numbers together from 1) and it is written as 10!.
This means there are 3,628,800 ways of arranging the 10 people

By theorem, The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ..., pk are of kth kind and the rest, if any, are of different kind is n!/(p₁! * p₂! *......* pk!)

Total flags = 4
Blue = 3
red = 1
Total no. of signals = (4!*3!)/3!
= 4

To begin with, we calculate the total number of possibilities that arise from tossing a coin 6 times . On each toss , we have 2 possibilities - a head or a tail. This gives us
2*2*2*2*2*2 = 64 possibilities
Now let's list out the desirable outcomes.
   Having 4 heads
H H H H T T - This is one example of the above outcome. Something like
H H H T H T would also be equally likely and would be a desirable outcome. Thus to calculate all such permutations
= (6!/4!) ∗ 2!
= 15ways

Option D is correct.

3! = 6 and 2! = 2.

So 3!/2! = 6 ÷ 2 = 3, which matches the right-hand side.

For completeness, 3! + 2! = 6 + 2 = 8, which is not equal to 5! = 120, and 3! × 2! = 6 × 2 = 12, which is not equal to 6! = 720. Also 1! = 1, so 3! + 2! ≠ 1!.

We have 5 choices (a, e, i, o, u)
3 letter codes = 5 * 5 * 5
= 125