Test: Basic Concepts Of Permutations And Combinations- 3 - CA Foundation MCQ

The vowels are never being separated.

So, vowels a & u are taken together and considered as one letter.

So, there are 6 letters in the word draught

The number of ways in which it can be arranged are given by =6!=720. 

But the internal arrangement can be done of a and u in 2 ways.

Hence the total arrangements are 720×2=1440.

ⁿP₄ represents the number of permutations of n things taken 4 at a time, and nP2 represents the number of permutations of n things taken 2 at a time.

We are given that ⁿP₄ = 20 (nP2).

Using the formula for permutations, we know that ⁿP₄ = n!/(n-4)!, and ⁿP₂ = n!/(n-2)!.

Substituting these expressions into the given equation, we get:

n!/(n-4)! = 20 * n!/(n-2)!

Cancelling out the common factor of n!/[(n-4)! * (n-2)!], we get:

(n-3) * (n-2) = 20

Expanding the left side of the equation, we get:

n² - 5n + 6 = 20

Simplifying, we get:

n² - 5n - 14 = 0

Factorizing, we get:

(n - 7) * (n + 2) = 0

Therefore, the solutions are n = 7 and n = -2.

However, n cannot be negative since we are counting permutations. Therefore, the only possible value of n is 7.

To include the red ball and exclude the blue ball, we need to choose 3 remaining balls from the 10 white balls. This can be done in 10C3 ways.

¹⁰C₃ = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = 10*9*8 / (3*2*1) = 120

So, the answer is ¹⁰C₃ or 120 ways. The correct option is:

¹⁰C₃

We have,1,2,3,4,5,6,7,8

Total counters = 8

Taken at a time = 4

The total combinations without restrictions= 8C4 = 70

Odd = 1,3,5,7

Even = 2,4,6,8

We have to find out those combinations which contain at least one even and one odd counter i.e. those neither can have all 4 even counters nor all 4 odd counters.

Let's first find the the no. of combinations with restrictions i.e. which contains all even or all odd counters.

So, the no. of combination which have all 4 odd counters = 4C4 = 1

Similarly, the no. of combination which have all 4 even counters = 4C4 = 1

Hence, the required no. combinations which contain at least one even and one odd counters = Total - Restricted combinations

= 70–1–1

= 68 Ans.

Given the word is

PROPORTION

When

Number of P is =2

Number of O is =3

Number of R is =2

Number of I is =1

Number of T is =1

Number of N is =1

Let us frist find the number of selections.

Case 1:- All four letters distinct = number of ways will be 4×6C4​=4!×4!(6−4)!6!​=2!6×5×4×3×2×1​=360

Case 2:- Three letters same and one letter distinct, then no. of ways  = 3!4!​5C1​=3!4×3!​×5=20

Case 3:- Two letters of one type and the other two letters of other type, then no. of ways(out for P,R and O)=3C2​×4C2​=3×2!×2!4!​=18

Case 4:-two letters same and other two letters are different.then, no. of ways. = 3C1​×5C2​=3×10=30×2!4!​=360

Now according to given question,

Total number of selections=15+5+3+30=53 

Total no. of balls =12

They said to exclude both red&blue
then, remaining total balls=10
out of this 4 should be selected
therefore, Req. answer=¹⁰c4

According to formula:

total no of ways to select questions|
= (3-1)! (4-1)!
= 2! 3!
= 2 x 3 x 2
= 12 ............. ans..

The word contain 11 letters with repetition of letters.
We can choose 4 letters by
i) all the four letters are distinct
ii) Two distinct and two alike
iii) Two alike of one kind and two alike of other kind
i) 8 different letters are M, A, T, H, E, I, C, S
This can be chosen by ⁸C₄ × 4! = 1680
ii) Three pairs alike letters MM, AA, TT. One pair can be chosen by ³C₁ = 3 ways 
Remaining two letters can be chosen by ⁷C₂ 
Each such groups has 4 letters out of which 2 are alike.
Hence, it can be chosen by 4!/2! 
Total number of ways = 3 × ⁷C₂ × 4!/2! = 756
iii) Three pairs of 2 alike letters. Two pairs can be chosen by ³C₂
 ³C₂ groups of 4 letters each.
There are 4 letters of which 2 alike of one kind and two alike of other kind in each group.
That can be arranged by 4!/2! × 1/2!
Total number of ways = ³C₂ × 4!/2! × 1/2! = 18
Total number of ways = 1680 + 756 + 18 = 2454

As there are 21 red balls and 19 blue balls.

So there can be 22 different positions for blue balls ( because we can't arrange two blue balls together).

Applying concept,

Total ways = 22C19

= 22!/19!(22-19)!

= 22!/19!*3!

= 22*21*20*19!/19!*3!

= 22*7*10

= 1540.