Test: Permutations and Combinations- 1 - CA Foundation MCQ

⁴P₃

= 4!/(4!-3!)

= (4!)/1!

= 4*3*2*!

= 24

0! = 1

n is a positive integer

• Recall that nPr = n! / (n-r)!. So, ⁿP₄ = n! / (n-4)! and ⁿP₂ = n! / (n-2)!
• Substituting these into the equation gives: n! / (n-4)! = 12 × (n! / (n-2)!)
• Simplifying leads to: (n(n-1))/(n-4) = 12
• This simplifies to: n(n-1) = 12(n-4)
• Expanding and solving yields n = 6.

Since all letters in the word "COMPUTER" are distinct then the arrangements is 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

Now we have 3 consonants and 4 Vowels. 
as all vowels have to come together let treat them one object so we now have 3 consonant and 1 this entity
so this can be arranged in 4! ways and 4 vowels internally in 4! ways so 
final answer would be 4!x4! = 24x24 = 576

No. of ways in which 10 paper can arrange is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10!- 9!.2! = 9!(10-2) = 8.9!.

N. particles arrange in a way = n!

if two particles are arranged together then
there is only (n-1) particles

when two particles are together = (n-1)!


and two particles are arranged in= 2! ways



Thus, number of ways in which no particles
are not come together


=n! -2×(n-1)!

=( n-1)!×(n-2)
 

To form a 4-digit number greater than 5000 using the digits 3, 4, 5, 6, and 7 without repetition, we need to consider the following:

1. The first digit has to be either 5, 6, or 7.
2. The remaining digits can be any of the remaining digits, i.e., 4 digits out of the 4 available.

For the first digit, there are 3 choices (5, 6, or 7).

Then, for the remaining three digits, we have 4 choices for the second digit, 3 choices for the third digit, and 2 choices for the fourth digit, since we cannot repeat any digit.

So, the total number of such 4-digit numbers is 3×4×3×2 =72.

There are 4 ways to pick the 1st digit (can't use 0 for the 1st digit).
There are 4 remaining ways to pick the 2nd digit. Can pick 0, but not the digit picked for the 1st digit.
There are 3 remaining ways to pick the 3rd digit.
There are 2 remaining ways to pick the 4th digit.
Answer: (4)(4)(3)(2) = 96 ways.

angle would be 1 word  so total letters would be 4 [ Tri + angle ] these 4 letters can be arranged in 4! ie 24 ways

There are only 3 vowels, a, u, and e, and there are 4 odd places.  So
one of the odd spaces must contain a consonant.
We can choose 3 of the 4 odd places to put the vowels in C(4,3) or 4 ways.
For each of those vowel , we can  rearrange them  in P(3,3) or 3! or 6 ways.
And we can rearrange the 5 consonants in P(5,5) or 5! or 120 ways.
Hence the total number of different words formed are 6*4*120 = 2880 

No. of ways in which n people can sit in a round table = (n-1)! 
So to make them sit together, consider them as a single unit in the sitting arrangement ie. n= 6.
Number of ways so that two particular boys may sit together in the round table:- 

(6 -1)!*2

(The two particular boys can also be switched among themselves) 

=> 120*2
=> 240 

No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!

Number of different things = 10

Numbers of things which are taken at a time = 4

Now, it is said that one particular thing always occur.

So The question reduces to selecting 9 things from the 3 things

This can be done easily by using the concept of combinations

Number of ways selecting 3 things out of 9 different things such that one particular thing always occur =

Now, the 4th thing is fixed so number of ways for selecting the 4th thing = 4! = 24

So, Required number of ways = 84 × 24 = 2016