**:: Questions ::**

**1. Find the values of**

- P(6, 6)

On comparing with P(n, r),

n = 6

r = 6

P(6, 6) = n! / (n - r)!

= 6! / (6 - 6)!

= 6! / 0!

= 720/1 ................. [since 0! = 1]

Therefore P(6, 6) = 720

- P(4, 2)

On comparing with P(n, r), we get

n = 4

r = 2

P(4, 2) = n! / (n - r)!

= 4!/ (4 - 2)!

= 4! / 2!

= 24/2

= 12

Therefore P(4, 2) = 12

- P(7, 2)

On comparing with P(n, r), we get

n = 7

r = 2

P(7, 2) = 7! / (7 - 2)!

= 7! / 5!

= 5040 / 120

= 42

Therefore the number of permutations is 42.

- P(70, 2)

On comparing with P(n, r), we get

n = 70

r = 2

Using the formula, P(n, r) = n! / (n - r)!

P(70, 2) = 70! / (70 - 2)!

= 70! / 68!

Now we know that 70! = 70 x 69 x 68!, therefore

= (70 x 69 x 68!) / 68!

= 70 x 69 ............... [the 68! from numerator is cancelled by 68! in the denominator]

= 4830

Therefore the number of permutations is 4830

**2. Find 'n' in the following equations:**

- P(n, 6) = P(n, 5)

Using the formula P(n, r) = n! / (n - r)!, we get

n! / (n - 6)! = n! / (n - 5)!

By cross multiplication, we get,

n! / (n - 6)! * (n - 5)! / n! = 1

(n! * (n - 5)!) / ((n - 6)! * n!) = 1

n! in the numerator is cancelled by n! in the denominator,

(n - 5)! / (n - 6)! = 1

We know that (n - 5)! = (n - 5)(n - 6)!, therefore

((n - 5)(n - 6)!) / (n - 6)! = 1

The (n - 6)! in the numerator is cancelled by the (n - 6)! in the denominator, therefore

(n - 5) = 1

Adding 5 to both sides, we get

n = 6

Therefore, the answer is n = 6

- P(n, 5) = P(n-1, 4)

Using the formula P(n, r) = n! / (n - r)!, we get

n! / (n - 5)! = (n - 1)! / (n - 1 - 4)!

n! / (n - 5)! = (n - 1)! / (n - 5)!

By cross multiplication, we get,

n! / (n - 5)! * (n - 5)! / (n - 1)! = 1

(n! * (n - 5)!) / ((n - 5)! * (n - 1)!) = 1

The (n - 5)! present in the numerator and denominator of the fraction above get cancelled out by each other, so we get

n! / (n - 1)! = 1

We know that n! = n(n - 1)!, so we get

n(n - 1)! / (n - 1)! = 1

The (n - 1)! in the numerator and denominator of the fraction above get cancelled out by each other, so we get

n = 1

Therefore, the value of n is 1.

**3. Fifteen students participate in a contest. In how many ways can the first three prizes be distributed?**

Total number of students, n = 15

Total number of prizes to be distributed, r = 3

Total number of permutations, P(n, r) = n! / (n - r)!

Therefore, P(15, 3) = 15! / (15 - 3)!

= 15! / 12!

We know that 15! = 15 x 14 x 13 x 12!, therefore

= 15 x 14 x 13 x 12! / 12!

The 12! present in the numerator and denominator get cancelled out, so we get,

= 15 x 14 x 13

= 2730

Therefore there are 2730 different ways in which 3 prizes can be distributed among 15 students.

**4. Find the number of signals which can be hosted using five flags of different color all at a time.**

The number of flags present, n = 5

Number of flags hosted at a time, r = 5

Therefore total number of permutations, P(n, r) = n! / (n - r)!

Therefore,

P(5, 5) = 5! / (5 - 5)!

P(5, 5) = 5! / 0! ................ [since o! = 1]

P(5, 5) = 5! = 120

Therefore the total number of ways in which all the 5 different flags can be hoisted are 120.

Hope the workings are clear. For any clarification, please post the problem in the comments. Thanks.

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