HirePro Permutations and Combination Quiz-1

The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 3000, the only 4-digit number to start with 3.
The leftmost digit (thousands place) of each of the 4 digit numbers other than 3000 can take one of the 2 values 1 or 2 .
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.
Hence, there are 2 x 5 x 5 x 5 or 250 numbers from 1000 to 2999.
Including 3000, there will be 251 such numbers.

The word “ACCESS” has 6 letters in which “C and S” comes twice.
If repetition is allowed, the number of words we can form = 4*4*4*4*4*4 = 4096.
(This is because, when repetition is allowed, we can put any of the four unique alphabets at each of the five positions.)
If repetition is not allowed, the number of words we can form =/frac{6!}{2!*2!}= 180.

The boys will be arranged in 17! ways. Now, there are a total of 18 possible places available between boys such that no 2 girls can be placed together (alternate sequence of boys and girls, starting and ending positions for girls).
Therefore, the 15 girls can stand at these 18 places only.
Hence, the number of ways = 17!* 18P15

A triangle is formed by joining any three non-collinear points in pairs.
There are total 15 point points ,for a triangle 3 point are required
= 15C3
5 collinear point which cannot make a triangle = 5C3
TOTAL TRIANGLES =15C3 - 5C3
= 445

There are 10 students and the maximum capacity of the cars together is 11.
We may divide the 10 students as follows
Case I: 6 students in the first car and 4 in the second
Case II: 5 students in the first car and 5 in the second
Hence, in Case I: 10 students are divided into groups of 6 and 4 in 10C4 ways.
Similarly, in Case II: 8 students are divided into two groups of 5 and 5 in 10C5 ways.
Therefore, the total number of ways in which 10 students can travel is:
10C4+10C5
=210 + 252= 462

A number is divisible by 5 if and only if its last digit is either 5 or 0. So, we have to fix 5 or 0 as the last digit of the 4-digit number and fill 3 places with the remaining 8 digits.
When last digit is 5
Number of ways to choose 3 digits = 7C3 = 35.
Number of ways to arrange the chosen digits = 3!
Total = 35*3!=210

There are 7 letters in the word PENGUIN of these 3 are vowels and 4 consonants.
There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4P3
ways and 4 constants can be arranged in /frac{4p4}{2!} ways.
Number of words =4P3 x /frac{4p4}{2!} =288

When E and U are the first and last letters of all the words then we can arrange them in 9!/2! ways. Similarly When U and E are the first and last letters of the words then the remaining letters can be arrange in 9!/2! ways.

Thus the total number of permutations = 2 x \frac{9!}{2!} =9!

As per the common explanation, we get
Total number of ways in which two girls are together = 2× 120= 240
Common Explanation: As per the question, two girls can’t occupy consecutive seat.

Therefore, if we find the number of ways in which all two girls occupy consecutive seats and subtract this number from the total number of ways in which the six people can be arranged among themselves, we will get the required answer.

Assume that the 2 girls are one entity. The total number of ways in which they can be arranged among themselves = 2! = 2

Also, the set of two girls and the other students can be arranged among themselves in 5! = 120 ways.

Thus, total number of ways in which two girls are together = 2 × 120 = 240

Thus, number of ways in which all 3 girls will not occupy consecutive seats = 720 – 240 = 480

Hence, the number of 4-letter words (with or without meaning) formed by using these letters = (12P4)

= 12 × 11 × 10 × 9 = 11880