The first application can be placed in 6 ways and the second application can be placed in 6 ways and so on.Since we have to place all the applications, the number of ways = 6 X 6 X 6 X 6 X 6 = (6 )⁵

The total number of ways in which five tourists can stay in eight hotels is (8)⁵ .The number of ways in which no two tourists stay together is 8 X 7 X 6 X 5 X 4 = 6720. Therefore, the number of ways in which at least two tourists stay together is = (8)⁵ – 6720 = 32768 - 6720 = 26048

Total number of four digit numbers is 9 X 10 X 10 X 10 = 9000.

Number of four digit numbers with number 5 in them can be found out in the following manner.

First digit can neither be zero nor 5. So, the first digit can be chosen in eight ways, the second,  third and fourth digit each can be chosen in 9 ways as there is no restriction of not choosing 0. Hence, the number of 4 digit numbers with number 5 in them = 8 X 9 X 9 X 9 = 5832.

Total number of four digit numbers with at least one 5 in them = 9000 – 5832 = 3168

Without repeating we can form 6 one digit numbers. 6 X 5 = 30, two digit numbers. 6 X 5 X 4 = 120, three digit numbers. 6 X 5 X 4 X 3 = 360, four digit numbers. 6 X 5 X 4 X 3 X 2 = 720, five digit numbers and 6 X 5 X 4 X 3 X 2 X 1 = 720 six digit numbers.

Since, this number formation in each case is mutually exclusive we will have to add them all up to arrive at the solution.

So according to problem we can have 6 + 30 + 120 + 360 + 720 + 720 = 1956 numbers.

In binary system only 0 and 1 are to be used and since 0 can not be used to fill the first place, so the number of 8 digit numbers in binary system are 1 X 2 X 2 X 2 X 2 X 2 X 2 X 2 = 128.

Since the dice are distinguishable, the number of ways one can obtain a sum of nine = (b3, g6), (b6, g3), (b4, g5),(b5, g4). As all of these ways occur separately, number of ways = 4.

There are three vowels, viz. E, O & U which are not to be separated. Let us assume them to be one letter EOU, then we have got 5 (consonants) + 1(vowels group) = 6 letters which can be arranged in 6 X 5 X 4 X 3 X 2 X 1 = 720 ways. But EOU themselves can be arranged in 3 X 2 X 1 = 6 ways, so total possible arrangements are 720 X 6 = 4320 ways.

We can make two letter, three letter and so on to six letter words. Except for two letter words, all other words will have letters filled between S and R. Therefore, the possible words thus are: SR = 1 , S_R = 4 , S_ _ R = 4 X 3 (since second place can have 4 and third place 3 possibilities) = 12, S_ _ _ R = 4 X 3 X 2 = 24, S_ _ _ _ R = 4 X 3 X 2 X 1 = 24. Therefore, total number of words = 1 + 4 + 12 + 24 + 24 = 65.

Since both N’s in NN@ can take any value from 0 to 9 except for one case where both are zero, so we can have 10 X 10 -1 = 99 such numbers and @ can take 26 different values.

In case of NNNN we can have 10 X 10 X 10 X 10 - 1 = 9999 possibilities since all N’s can not be zero.

Therefore, maximum such number plates possible = 99 X 26 X 9999 = 25737426.

We have 1 blue,1 yellow, 3 green, 2 white, 1purple and 1 violet ball. That makes a total of 9 balls which can be arranged in 9! ways but since 3 green and 2 white balls are indistinguishable and themselves can be arranged in 3! and 2! ways respectively, so we will have to divide them with total to negate repetitions.

Therefore, total number of ways = (9!)/(3!2!) = 30240.

Let there be P persons in the party. A handshake takes place every time two persons are chosen out of these P persons. Therefore, total number of handshakes = Total ways of choosing 2 persons out of P = ^PC₂ = P! / 2!(P - 2)! = P(P – 1)/2 = 21 or P = 7.

For the first question 5 answers are possible, for the second question, there also 5 answers are possible, … and so on. Therefore, the total number of possible answer keys = 5 X 5 X 5 …(fifteen times) = (5)¹⁵