Since the total members are 15, but one is to the left, because of circular condition. Therefore, the remaining members are 14, but three special members constitute a member. Therefore the required number of arrangements are 12 ! ×2, because the chairman remains between the two specified persons and the person can sit in two ways.

p = 10, q = 9, r = 7

Total ways of selection = (p + 1) * (q + 1) * (r + 1) – 1

= [11 * 10 * 8] – 1

= 879

Since the total number of ways in which boys can occupy any place is (5 − 1)! = 4 ! and the 5 girls can sit accordingly in 5 ! ways. Hence the required number of ways is 4 ! ×5 !.

Since, 38808 = 8 × 4851

= 8 × 9 × 539

= 8 × 9 × 7 × 7 × 11

= 2³ × 3² × 7² × 11

So, the number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72.

This includes two divisors 1 and 38808.

Hence, the required number of divisors = 72 – 2 = 70.

We have,

x* y * z = 24

x* y * z = 2³ × 3¹

The number of ways of distributing ‘n’ identical balls into ‘r’ different boxes is ^{(n + r − 1)}C_{(r − 1)}

Here we have to group 4 numbers into three groups

Number of integral positive solutions

= ^{(3 + 3 − 1)}C_{(3 − 1)} × ^{(1+ 3 − 1)}C_{(3 − 1)}

= ⁵C₂ × ³C₂

= 30

Case I: When only one flag is used. No. of signals made = ⁵P₁ = 5.

Case II: When only two flags are used. Number of signals made = ⁵P₂ = 5 * 4 = 20.

Case III: When only three flags are used. Number of signals is made = ⁵P₃ = 5 * 4 * 3 = 60.

Case IV : When only four flags are used. Number of signals made = ⁵P₄ = 5 * 4 * 3 * 2 = 120.

Case V : When five flags are used. Number of signals made = ⁵P₅ = 5! = 120.

Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.

Since at any place, any of the digits 2, 5 and 7 can be used, the total number of such positive n-digit numbers are 3ⁿ. Since we have to form 900 distinct numbers, hence 3ⁿ ≥ 900 ⇒ n = 7.

4 novels can be selected from 6 novels in ⁶C₄ ways. 1 dictionary can be selected from 3 dictionaries in ³C₁ ways. As the dictionary selected is fixed in the middle, the remaining 4 novels can be arranged in 4! ways.

Hence, the required number of ways of arrangement = ⁶C₄ * ³C_1.* 4! = 1080

We know that a five-digit number is divisible by 3, if and only if the sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five-digit numbers.

Now,

(i) In this case, we do not use 0; the five-digit number can be formed (from the digit 1, 2, 3, 4, 5) in ⁵P₅ ways.

(ii) In this case, we do not use 3; the five-digit number can be formed (from the digit 0, 1, 2, 4, 5) in 6 ! −5! × 2 = 480 ways.

The total number of such 5 digit number = ⁵P₅ + (⁵P₅ − ⁴P₄)

= 120 + 96

= 216.

Two rings may have the same letter at a time, but the same ring cannot have two letters at a time. Therefore, we must proceed ring wise. Each of the three rings can have any one of the 10 different letters in 10 ways.

Therefore, the total number of attempts = 10 × 10 × 10 = 1000.

But out of these 1000 attempts, only one attempt is successful.

Therefore, the required number of unsuccessful attempts = 1000 – 1 = 999.