CAT Test: Permutation & Combination- 5 Free Online Test 2026

A and B can occupy the first and the ninth places, the second and the tenth places, the third and the eleventh place and so on... This can be done in 18 ways.
A and B can be arranged in 2 ways.
All the other 24 alphabets can be arranged in 24! ways.
Hence the required answer = 2 x 18 x 24!

The numbers between 200 and 1200 can be formed in the following ways:

1) Numbers with 3 digits: The hundreds place can be filled with 2 or 3, so there are 2 ways. The tens place can be filled with any of the remaining 3 digits, so there are 3 ways. The units place can be filled with any of the remaining 2 digits, so there are 2 ways. Therefore, there are 2*3*2 = 12 such numbers.

2) Numbers with 4 digits: The thousands place can only be filled with 1, so there is 1 way. The hundreds place can be filled with any of the remaining 3 digits (not including 0), so there are 3 ways. The tens place can be filled with any of the remaining 2 digits, so there are 2 ways. The units place can be filled with the remaining 1 digit, so there is 1 way. Therefore, there are 1*3*2*1 = 6 such numbers.

Therefore, a total of 12 + 6 = 18 numbers between 200 and 1200 can be formed with the digits 0, 1, 2, 3 with no repetition of digits. So, option 2 is correct.

²⁸C_2r/²⁴C_2r-4 = 225/11
⇒ 28!/2r!(28-2r)! * [(2r-4)!(28-2r)!]/24! = 225/11
⇒ (28*27*26*25)/[2r*(2r-1)*(2r-2)*(2r-3)] = 225/11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = (28*27*26*25*11)/225
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = 28*3*26*11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = 4*7*3*13*2*11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = (2*7)*13*(3*4)*11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = 14*13*12*11
⇒ 2r = 14
⇒ r = 7

We have to count natural numbers which have a maximum of 4 digits. The required answer will be given by: Number of single digit numbers + Number of two digit numbers + Number of three digit numbers + Number of four digit numbers.

Explaination: out of 11 player 1 captain can be choose 11 ways, Now remaining 10 player,wise captain can be choose in 10 ways Therefore total number of ways =11*10=110 ways

In the ALLAHABAD :-
only 1 vowel available for selection (A).
A is available 4 times .
there are 4 consonants available – L, H, B, D
Then the number of ways of selecting a vowel and a consonant would be 1 × ⁴C₁ = 4.

The correct option is Option D.

Since motor registration can start with zero as first digit from left and there are six different digits (0,1,2,3,4,5) including zero.

Therefore, different registration numbers that can be formed using these six digits = ⁶P₄  = 360

There are 10 subjects and 5 periods
I st period can be filled with any 10 subjects
2nd period can be filled with remaining 9 subjects (1 subject is already filled)
3 rd period can be filled with remaining 8 subjects (2 subjects are already filled)
4 th period can be filled with remaining 7 subjects (3 subjects are already filled)
5 rd period can be filled with remaining 6 subjects (4 subjects are already filled)
so total no fo ways we can arrange = 10*9*8*7*6
= 30240 ways

The selection of the II player team can be done in ¹⁴C₁₀ ways. This results in the team of 11 players be-ing completely chosen. The arrangements of these 11 players can be done in 11!.
Total batting orders = ¹⁴C₁₀ x 11! = 1001 x 11!
(Note: Arrangement is required here because we are talking about forming batting orders). 

The word PROWLING consists of 8 letters, 2 of which are to be fixed (R at the beginning and W at the end).
This leaves us with 6 letters (P, O, L, I, N, G) to arrange in the remaining 6 places, and all of these letters are different.

The number of ways to arrange n different items is given by n factorial (n!).

Therefore, the number of distinct words that can be formed is 6! = 720.

for an even number, the ones place number should be even i.e it could be either one of 2,4,6

so number of ways to select ones place number is ³C₁ = 3

since repetetion is allowed, and all numbers are whole numbers.

so number of ways to select any of the three remaining numbers = ⁶C₁ =6

Number of 4 digit number possible = 6 x 6 x 6 x 3

= 648

First, we consider each subject as one group. So, we have 3 groups to arrange. The number of ways to arrange these groups is 3!.
Within each group, the books can also be arranged. The number of ways to arrange 2 Geology books is 2!, the number of ways to arrange 2 Sociology books is 2!, and the number of ways to arrange 5 Economics books is 5!.
So the total number of ways to arrange the books is 3! * 2! * 2! * 5! = 6 * 2 * 2 * 120 = 2880 ways.

The correct option is Option B.

The word equation has 8 letters of which 5 are vowels and 3 are consonants. Bunch up the 5 vowels to assume them as one letter. Thus the total number of letters is 4 which can be arranged in  4!=24 . And for each of these arrangements the 5 vowels can be arranged among themselves in  5!=120  ways. Thus, the total arrangements of the letters with the vowels always appearing together is  24∗120=2880 .

otal number of quadrilateral combination possible if none of the points are collinear = ²⁵C₄ = 12650

If we form a geometry by joining any three points out of seven collinear points and one point from 18 non collinear points, it will give us a triangle instead of quadrilateral. So we have to eliminate number of combinations which can be formed in this way, which is ⁷C₃ x ¹⁸C₁ = 35 x 18 = 630

We also can't form quadrilateral if we choose all four vertices of quadrilateral to be any 4 points from 7 collinear points. It will come out to be a straight line. So we have to eliminate such combinations also. Which is ⁷C₄ = 35

So net number of possible quadrilaterals = 12650 - 630 - 35 = 11985

Choose 5 men out of 9 men = ⁹C₅ ways = 126 ways

Choose 3 women out of 12 women = ¹²C₃ ways = 220 ways

The committee can be chosen in 27720 ways