Permutation-Combination Exercise 2

Required number of ways = ways of selecting 4 objects out of 6 given objects

= ⁶C₄ = (6 × 5)/2 = 15

It is a question of arrangement without repetitions.

Required no. of ways = 5 × 4 × 3 = 60

First of all we will prime factorize 8064.

8064 = 2 × 4032

= 2² × 2016

= 2³ × 1008 = 2⁴ × 504

= 2⁵ × 252 = 2⁶ × 126 = 2⁷ × 63

= 2⁷ × 3² × 7¹

Required no. of ways

= (7 + 1) (2 + 1) . 1

= 8 × 3 = 24

Two particular girls can be arranged in 2! ways and remaining 10 girls can be arranged in 10! ways.
Required no. of ways = 2! × 10!

5 letters out of 15 letters can be selected in ¹⁵C₅ ways.

The thousandth place can be filled up in 9 ways with any one of the digits 1, 2, 3, ...., 9.

After that the other three places can be filled up in ⁹P₃ ways, with any one of the remaining 9 digits including zero.

Hence, the number of four digit numbers with distinct digits = 9 × ⁹P₃

The first and the last (terminal) digits are even and there are three even digits.

This arrangement can be done in ³P₂ ways.

For any one of these arrangements, two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits (leaving out the terminal digits) may be arranged using these 5 digits in ⁵P₄ ways.

The required number of numbers is

³P₂ × ⁵P₄ = 6 × 120 = 720.

Any number greater than a million must be of 7 or more than 7 digits.

Here number of given digits is seven, therefore we have to form numbers of seven digits only.

Now there are seven digits of which 3 occurs thrice and two occurs twice.

∴ number of numbers formed

= = 420

But this also includes those numbers of seven digits whose first digit is zero and so in fact they are only six digit numbers.

Number of numbers of seven digits having zero in the first place = 60.

Hence required number

= 420 – 60 = 360

Required number of numbers

= 3 × 5 × 5 × 5 = 375

Required sum

= (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ...+ 100) – (10 + 20 + ... + 100)

= 2550 + 1050 – 530 = 3070.

X - X - X - X - X.

The four digits 3, 3, 5,5 can be arranged at (-) places in

= 6 ways.

The five digits 2, 2, 8, 8, 8 can be arranged at (X) places in

ways = 10 ways

Total no. of arrangements

= 6 × 10 = 60 ways

⇒

⇒

⇒

⇒

⇒

⇒ 10 - r = 5

⇒ r = 5

The inequality is

We must have n + 1 ≥ 3 and n + 1 ≥ 2

⇒ n ≥ 2 and n ≥ 1

⇒ n ≥ 2 and also

⇒ (n + 1) n(n - 4) ≤ 600

By trial the values of n satisfying this are 2, 3, 4, 5, 6, 7, 8, 9 which are eight in number.

⇒ m > 6

The least value of m is 7.

⇒ r² = 3r

⇒ r² = 40 - 3r

⇒ r =0, 3 or -8, 5

3 and 5 are the values as the given equation is not defined by r = 0 and r = –8.

Hence, the number of values of r is 2.

We have, ⁿP_r = ⁿP_r+1

⇒ n – r = 1 ...(1)

Also, ⁿC_r = ⁿC_r–1

⇒ r + r – 1 = n

⇒ 2r – n = 1 ...(2)

Solving (1) and (2), we get r = 2 and n = 3

ⁿP_r = 720ⁿC_r

⇒

⇒ r! = 720 = 1 × 2 × 3× 4 × 5 × 6!

⇒ r = 6

We have

=

=

=

⇒ m = 10, n = 12

Given, kP₃ = 9240

⇒ k(k - 1) (k - 2) = 9240 = 22 × 21 × 20

⇒ k = 22

∴ m + n – k = 10 + 12 – 22 = 0

Given,

=

..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....

⇒ r² - r - 132 = 0

⇒ (r - 12) (r + 11) = 0

⇒ r = 12

⇒ n ≥ 12

So, minimum value of n = 12.

Number of all possible triangles

= Number of selections of 3 points from 8 vertices

= ⁸C₃ = 56

Number of triangle with one side common with octagon

= 8 × 4 = 32

(Consider side A₁A₂. Since two points A₃, A₈ are adjacent, 3rd point should be chosen from remaining 4 points.)

Number of triangles having two sides common with octagon : All such triangles have three consecutive vertices, viz., A₁A₂A₃, A₂A₃A₄, ..... A₈A₁A₂.

Number of such triangles = 8

∴ Number of triangles with no side common

= 56 – 32 – 8 = 16.