We know that, each AP consists of a unique first term and a common difference.

So, number of ways to select the first term of a given set is ^{3}C_{1} = 3

And, number of ways to select a common difference of given set is ^{5}C_{1} = 5

Hence, total number of AP’s possible are ^{3}C_{1} × ^{5}C_{1} = 3 × 5 = 15

How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’, assuming that no letter is repeated, if

**(i)** 4 letters are used at a time

**(ii)** all letters are used at a time

**(iii)** all letters are used but first letter is a vowel ?

There are 10 persons named P_{1}, P_{2}, P_{3} …, P_{10}. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement P_{1} must occur whereas P_{4} and P_{5} do not occur. Find the number of such possible arrangements.

How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?