One ball for box 3 can be selected in ⁷C₁ ways.
Two balls for box 5 can be selected in ⁶C₂ ways.

Remaining balls = 4
Remaining boxes = 3

In these 4 balls, 1^st ball can be put in any of these 3 boxes.
Similarly 2^nd ball can be put in any of these 3 boxes.
3^rd ball can be put in any of these 3 boxes.
4^th ball can be put in any of these 3 boxes.

i.e., these 4 balls can be arranged in
$3\times 3\times 3\times 3=3^4$ ways

Required number of ways
= ⁷C₁ × ⁶C₂ $\times 3^4$
$=7\times 15\times 81=8505$

Case 1: 1 box gets 3 balls, the other 4 boxes each get 1 ball
Number of box options for the box with 3 balls = 5. (Any of the 5 boxes.)
From the 7 balls, the number of ways to choose 3 balls for this box = 7C3 = (7*6*5)/(3*2*1) = 35.
Number of ways to arrange the remaining 4 balls = 4! = 24.
To combine these options, we multiply:
5*35*24 = 4200.

Case 2: 2 boxes each get 2 balls, the other 3 boxes each get 1 ball
From the 5 boxes, the number of ways to choose a pair of boxes to hold 2 balls each = 7C2 = (5*4)/(2*1) = 10.
From the 7 balls, the number of ways to choose a pair of balls for the 1st box = (7*6)/*2*1) = 21.
From the remaining 5 balls, the number of ways to choose a pair of balls for the 2nd box = (5*4)/(2*1) = 10.
Number of ways to arrange the remaining 3 balls = 3! = 6.
To combine these options, we multiply:
10*21*10*6 = 12600.

Total ways = 4200+12600 = 16800.

Number of ways in which one or more objects can be selected from n distinct objects (i.e., we can select 1 or 2 or 3 or … or n objects at a time)
= ⁿC₁ + ⁿC₂ + … + ⁿC_n = 2ⁿ – 1

It is explicitly stated that 12 black balls are different, 7 red balls are different and 6 blue balls are different. Hence there are 25 (=12+ 7+ 6) different balls.

We can select one ball from 25 balls, two balls from 25 balls, … 25 balls from 25 balls.

Hence, required number of ways.

= Number of ways in which 1 ball can be selected from 25 distinct balls
+ Number of ways in which 2 balls can be selected from 25 distinct balls
+ Number of ways in which 3 balls can be selected from 25 distinct balls
…
+ Number of ways in which 25 balls can be selected from 25 distinct balls

= ²⁵C₁ + ²⁵C₂ + … + ²⁵C₂₅
$=2^{25}-1$

Here, no two consecutive digits can be the same.

The ten thousands place can be filled by any of the 9 digits (1,2,3,4,… 9)

Repetition is allowed here. Only restriction is that no two consecutive digits can be the same. Hence the digit we placed in the ten thousands place cannot be used at the thousands place. Hence thousands place can be filled by any of the 8 digits.

Similarly, hundreds place, tens place and unit place can be filled by any of the 8 digits

Hence, the required count of 5 digit numbers that can be formed using the digits 1,2,3,4,… 9 such that no two consecutive digits are same
$=9\times 8^4$

Given that any number of flags can be hoisted at a time. Hence we need to find out number of signals that can be made using 1 flag, 2 flags, 3 flags, 4 flags, 5 flags and 6 flags and then add all these.

Number of signals that can be made using 1 flag
= ⁶P₁ $=6$

Number of signals that can be made using 2 flags
= ⁶P₂ $=6\times 5=30$

Number of signals that can be made using 3 flags
= ⁶P₃ $=6\times 5\times 4=120$

Number of signals that can be made using 4 flags
= ⁶P₄ $=6\times 5\times 4\times 3=360$

Number of signals that can be made using 5 flags
= ⁶P₅ $=6\times 5\times 4\times 3\times 2=720$

Number of signals that can be made using 6 flags
= ⁶P₆ $=6\times 5\times 4\times 3\times 2\times 1=720$

Therefore, required number of signals
$=6+30+120+360+720+720=1956$

10 civil engineers can be arranged around a round table in
$(10-1)!=9!$ ways …(A)

Now we need to arrange software engineers such that software engineers and civil engineers are seated alternatively. i.e., we can arrange 10 software engineers in the 10 positions marked as * as shown below.

This can be done in $10!$ ways …(B)

From (A) and (B),
required number of ways $=9!\times 10!$

8 black balls can be arranged in 8! ways …(A)

Now we need to arrange white balls such that white balls and black balls are positioned alternatively. i.e., we can arrange 8 white balls either in the 8 positions marked as A,B,C,D,E,F,G,H or in the 8 positions marked as B,C,D,E,F,G,H,I as shown below.

8 white balls can be arranged in the 8 positions marked as A,B,C,D,E,F,G,H in 8! ways.

8 white balls can be arranged in the 8 positions marked as B,C,D,E,F,G,H,I in 8! ways.

8 white balls can be arranged in the 8 positions marked as A,B,C,D,E,F,G,H or in the 8 positions marked as B,C,D,E,F,G,H,I in $8!+8!=2\times 8!$ ways …(B)

From (A) and (B),
required number of ways $=8!\times 2\times 8!=2\times (8!{)}^2$

Total number of ways in which 9 different colour balls can be arranged in a row
$=9!\cdots$(A)

Now we will find out total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are always together.

We have total 9 balls. Since black, white, red and green balls are always together, group these 4 balls together and consider as a single ball. Hence we can take total number of balls as 6. These 6 balls can be arranged in $6!$ ways.

We had grouped 4 balls together. These 4 balls can be arranged among themselves in $4!$ways.

Hence, total number of ways in which 9 different colour balls be arranged in a row so that black, white, red and green balls are always together
$=6!\times 4!\cdots$(B)

From (A) and (B),
Total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are never together
$=9!–6!\times 4!=6!\times 7\times 8\times 9-6!\times 4!=6!(7\times 8\times 9–4!)=6!(504–24)=6!\times 480=720\times 480=345600$

1 book: 5 volume
1 book: 5 volume
1 book: 2 volume
1 book: 2 volume

Given that volumes of the same book should remain together. Hence, just tie the same volume books together and consider as a single book. Hence we can take total number of books as 4. These 4 books can be arranged in ⁴P₄ $=4!$ ways.

5 volumes of the 1^st book can be arranged among themselves in ⁵P₅ $=5!$ ways.

5 volumes of the 2^st book can be arranged among themselves in ⁵P₅ $=5!$ ways.

2 volumes of the 3^rd book can be arranged among themselves in ²P₂ $=2!$ ways.

2 volumes of the 4^th book can be arranged among themselves in ²P₂ $=2!$ ways.

Hence total number of ways
$=4!\times 5!\times 5!\times 2!\times 2!$

Total number of ways in which we can arrange 10 books on a shelf
= ¹⁰P₁₀ $=10!\cdots$ (A)

Now we will find out total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together.

We have a total of 10 books. If a particular pair of books must always be together, just tie these two books together and consider as a single book. Hence we can take total number of books as 9. These 9 books can be arranged in ⁹P₉ $=9!$ ways.

We had tied two books together. These books can be arranged among themselves in ²P₂ $=2!$ ways.

Hence, total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together
$=9!\times 2!\cdots$ (B)

From (A) and (B),
Total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will never be together

10!–(9!×2!)=10!–(9!×2)=(9!×10)−(9!×2)=9!(10−2)=9!×8

Suppose there are n points in a plane out of which m points are collinear. Number of triangles that can be formed by joining these n points as vertices
= ⁿC₃ – ^mC₃

10 students can be arranged in a row in ¹⁰P₁₀ = 10! ways.