Permutation, Combination and Probability (UPSC) Part-I

hese can be two cases :
Case I - One box contain 3 balls and rest two Contains 1 ball each.
Case II - One box contain 1 ball and rest two Contains 2 balls each.

Case 1 : Number of ways = ⁵C₃ × ²C₁ × ¹C₁ = 20

Now, these 3 boxes can be arranged in

3!/2!​

among themselves,

as two of them contains similar number of balls.
So, number of ways = 20 × 3!/2!​ = 60

Case II : Number of ways = ⁵C₁ × ⁴C₂ × ²C₂ = 30

Also, these 3 boxes can be arranged among themselves in

3!/2!​

, as two of them contains similar number of balls.

Thus, number of ways = 30 ×

3!/2!​ = 90

Now, either of case (I) or case (II) is possible.
Hence, total number of ways = 60 + 90 = 150

There can be 3 cases :
I. When one dice shows 2.
II. When two dices show 2.
III. When three dices show 2.

Case I : The dice which shows 2 can be selected out of the 3 dices in ³C₁ ways.
Remaining 2 dices can have any 5 numbers except 2. So number of ways for them = ⁵C₁ each, so no. of ways when one dice shows 2 = ³C₁ × ⁵C₁ × ⁵C₁.

Case II : Two dices, showing 2 can be selected out of the 3 dices in ³C₂ ways and the rest one can have any 5 numbers except 2, so number of ways for the remaining 1 dice = 5.
So, number of ways, when two dices show 2 = ³C₂ × 5

Case III : When three dices show 2 then these can be selected in ³C₃ ways.
So, number of ways, when three dices show 2 = ³C₃ = 1

As, either of these three cases are possible.
Hence, total number of ways = (3 × 5 × 5) + (3 × 5) + 1 = 91

3 balls can be placed in any of the 12 squares in ¹²C₃ ways.
Total number of arrangements = ¹²C₃ = 220
Now, assume that balls lie along the same line.
There can be 3 cases :

Case I : When balls lie along the straight horizontal line.
3 balls can be put in any of the 4 boxes along the horizontal row in ⁴C₃ ways.
Now, since there are 3 rows, so number of ways for case I = ⁴C₃ × 3 = 12

Case II : When balls lie along the vertical straight line 3 balls can be put in any of the 3 boxes along the vertical row in ³C₃ ways.
Now, as there are 4 vertical rows, so number of ways for Case II = ³C₃ × 4 = 4

Case III : Balls lie along the 2 diagonal lines towards the left and 2 diagonal lines towards the right.
Number of ways = 2 + 2 = 4

Number of ways, when balls lie along the line = 12 + 4 + 4 = 20

Number of ways when balls don't lie along the line = Total number of ways - number of ways when balls lie along the line.
= 220 - 20 = 200

Let us take books A and B as one i.e., they are always continuous.
Now, number of books = 4 - 2 + 1 = 3
These three books can be arranged in 3! ways and also A and B can be arranged in 2 ways among themselves.
So, number of ways when books A and B are always continuous = 2 × 3!

Total number of ways of arrangement of A, B, C and D = 4!
Hence, number of ways when A and B are never continuous =
Total number of ways - number of ways when A and B always continuous = 4! - 2 × 3! = 12

Suppose any particular student is always selected.
Now, remaining 2 students are to be selected out of the remaining 5 students.

It can be done in ⁵C₂ ways =

5!/2! × 3! = 10

There can be two cases :
Case I : When 1 row contains 3 balls and rest two contains 1 ball each.
Now, the row which contains 3 balls can be selected out of the 3 rows in ³C₁ ways and in this row number of ways of arrangement = ³C₃. In other two rows, number of ways of arrangement in each = ³C₁.
Thus, number of ways for case I = ³C₁ × ³C₃ × ³C₁ × ³C₁
= 3 × 1 × 3 × 3 = 27

Case II : When 1 row contains 1 ball and rest two rows contain 2 balls each.
This row, containing 1 ball can be selected in ³C₁ ways and number of ways of arrangement in this row = ³C₁. In other two rows, containing 2 balls each, number of ways of arrangement in each = ³C₂.
Thus, number of ways for case II = ³C₁ × ³C₁ × ³C₂ × ³C₂ = 3 × 3 × 3 × 3 = 81

As, either of these two cases are possible, hence total number of ways = case I or case II = 27 + 81 = 108.