CAT Probability, Combinatorics Questions

Let's take the number of paths between Q and R to be b and the number of paths between R and S to be a

We are given the paths from P to S through R (which would be 4a), the paths from P to S through Q (which would be 12) and the paths from P to Q to R to S, which would be 3ab) is equal to 62
Giving the relation 4a+12+3ab = 62
Or 4a+3ab = 50

The paths from Q to R directly (which would be b), through P( which would be 12) and through S (which would be 4a) are 27
Giving the relation b+12+4a = 27
Or 4a+b = 15

Subtracting this equation from the first one we got, we get 3ab-b=35, or b(3a-1)=35
b can be 1, 3, 5 or 7
Substituting these values in the second equation, we see that it can not be 1 or 5, leaving only 3 or 7 as the possible values. 

Substituting b as 3 in the first equation would give 13a=50, which is not true. 
Substituting bas 7 in the first equation would give 25a= 50, which would give a=2

We are asked the number of paths from Q to R, which is b=7

Therefore, 7 is the correct answer. 

Let the number of balloons each child received be 2a, 2b, 2c and 2d

2a + 2b + 2c + 2d = 20

a + b + c + d = 10

Each of them should get more than zero balloons.

Therefore, total number of ways = $$(n-1)_{C_{r-1}}=(10-1)_{C_{4-1}}=9_{C_3}=84$$

The number of 4-digit numbers possible using 1,1,2, and 4 is $$\frac{4!}{2!}=12$$

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean = $$\frac{24\left(1111\right)}{12}=2222$$

The answer is option A.

Case 1: 4-digit numbers

Given digits - 0, 1, 2, 3, 4, 5

_, _, _, _

As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.

For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways

Total number of possible 4-digit numbers = 4*60 = 240

Case 2: 5-digit numbers

_, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2 = 600

Case 3: 6-digit numbers

_, _, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2*1 = 600

Total number of integers possible = 600 + 600 + 240 = 1440

The answer is option A.

This question is an application of the product rule in probability and combinatorics.

In the product rule, if two events A and B can occur in x and y ways, and for an event E, both events A and B need to take place, the number of ways that E can occur is xy. This can be expanded to 3 or more events as well.

Event 1: Distribution of balloons

Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them. 

So we are left with 15 - 4 x 3 = 15 - 12 = 3 balloons and 3 children. 

Now we need to distribute 3 identical balloons to 3 children. 

This can be done in $$^{n+r-1}C_{r-1}$$ ways, where n = 3 and r = 3. 

So, number of ways = $$^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$$

Event 2: Distribution of pencils

Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with 6 - 3 = 3 pencils.

We now need to distribute 3 identical pencils to 3 children.

This can be done in $$^{n+r-1}C_{r-1}$$ ways, where n = 3 and r = 3.

So, number of ways = $$^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$$

Event 3: Distribution of erasers

We need to distribute 3 identical erasers to 3 children.

This can be done in $$^{n+r-1}C_{r-1}$$ ways, where n = 3 and r = 3.

So, number of ways = $$^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$$

Applying the product rule, we get the total number of ways = 10 x 10 x 10 = 1000.

The possible arrangements are of the form 

35 _ Can be chosen in 6 ways.

35 _ _ We can choose 2 out of the remaining 6 in $$^6C_2=15$$ways. We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in $$^6C_3=20$$ways. We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in $$^6C_4=15$$ways. We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.

= 16 possibilities

But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 - 1 = 15 cases.

Total = 16+15+16 = 47 possibilities

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.

Then (100c+10b+a)-(100a+10b+c)=198

99c-99a=198

c-a = 2.

Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.

Thus, there can be 7 cases.

B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be $$7\times\ 10=70$$.

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$$\frac{4!}{3!}$$ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$$\frac{4!}{2!\cdot2!}=\ 6$$

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$$\frac{4!}{3!}$$ = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively


  We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once 
So the possible cases :

Now we get $$\frac{4!}{2!}\times\ 3$$= 36 ( When one digit is used twice and the remaining two once )
$$\frac{4!}{3!}\times\ 2$$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$$\frac{4!}{2!\times\ 2!}=\ 6$$( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = $$^{(3+5)}C_3$$ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = $$^{(4+4)}C_4$$ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

Let 'ab' be the two digit number. Where b $$\neq$$ 0.

We will get number 'ba' after interchanging its digit. 

It is given that 10a+b > 3*(10b + a)

7a > 29b

If b = 1, then a = {5, 6, 7, 8, 9}

If b = 2, then a = {9}

If b = 3, then no value of 'a' is possible. Hence, we can say that there are a total of 6 such numbers.

In a tournament, there are 43 junior level and 51 senior level participants.

Let 'n' be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.

$$\Rightarrow$$ nC2 = 153

$$\Rightarrow$$ n(n-1)/2 = 153

$$\Rightarrow$$ n(n-1) = 306

=> n$$^{2}$$-n-306 = 0

=> (n+17)(n-18)=0

=> n=18  (rejecting n=-17)

Therefore, number of boys on junior level = 43 - 18 = 25. 

Let 'm' be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.

$$\Rightarrow$$ mC2 = 276

$$\Rightarrow$$ m = 24

Therefore, number of girls on senior level = 51 - 24 = 27. 

Hence, the number of matches a boy plays against a girl  = 18*25+24*27 = 1098

It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number.
There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $$9C_2$$ ways.
3 digit numbers can be formed in $$9C_3$$ ways.
..................................................
..9 digit number can be formed in 9C9 ways. 

We know that $$nC_0+nC_1+nC_2+.........+nC_n =2^n$$
=> $$9C_0 + 9C_1+9C_2+....9C_9 = 2^9$$
$$9C_0 + 9C_1 + ...9C_9 = 512$$

We have to subtract $$9C_0$$ and $$9C_1$$ from both the sides of the equations since we cannot form single digit numbers. 
=> $$9C_2 + 9C_3+...+9C_9=512-1-9$$
$$9C_2+9C_3+...+9C_9=502$$

Therefore, $$502$$ is the right answer. 

According to the question each column, each row, and each of the two diagonals of the 3X3 matrix add up to the same value. This value must be 15.

Let us consider the matrix as shown below: 

Now we'll try substituting values from 1 to 9 in the exact middle grid shown as 'x'.

If x = 1 or 3, then the value in the left bottom grid will be more than 9 which is not possible.

x cannot be equal to 2.

If x = 4, value in the left bottom grid will be 9. But then addition of first column will come out to be more than 15. Hence, not possible.

If x=5, we get the grid as shown below:

Hence, for x = 5 all conditions are satisfied. We see that the bottom middle entry is 3. 

Hence, 3 is the correct answer. 

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens
Now these two pens have to be divided between three persons so that each person can get zero pens = $$^{2+3-1}C_{3-1}$$ = $$^4C_2$$  = 6

The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as

Case I: 2, 3, 4, 6
Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways.
Hence total number of ways = 3*6 = 18

Case II: 0, 2, 3, 4
case II a: 0 is in the units place => 3! = 6 ways
case II b: 0 is not in the units place => units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 * 2 * 2 = 8
Total number of ways in this case = 6 + 8 = 14 ways.

Case III: 0, 2, 4, 6
case III a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 * 2 * 2 = 12
Total number of ways in this case = 6 + 12= 18 ways.

Hence the total number of ways = 18 + 14  + 18 = 50 ways

We have been given that a + b + c + d = 7
Total ways of distributing 7 things among 4 people so that each one gets at least one = $$^{n-1}C_{r-1}$$ = 6C3 = 20
Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 - 4 = 16

We have to essentially look at numbers between 1000 and 4000 (including both).

The first digit can be either 1 or 2 or 3.

The second digit can be any of the five numbers.

The third digit can be any of the five numbers.

The fourth digit can also be any of the five numbers.

So, total is 3*5*5*5 = 375.

However, we have ignored the number 4000 in this calculation and hence the total is 375+1=376

The power is 20.

20 has to be divided among a, b and c. This can be done in $$^{20+3-1}C_{3-1}$$ = $$^{22}C_2$$ = 231

Option a) is the correct answer.

The shortest route from A to B is via the diagonal OQ in the square P. One can travel from A to O in 4!/2!*2! ways. The shortest way from O to Q is through the diagonal only.From Q to B can be travelled in 6!/4!*2! ways.

The total number of ways is, therefore, (4!/2!*2!) * (6!/4!*2!) = 6*15 = 90

The shortest route from A to B is via the diagonal in the square P. A to the north-west corner of square P can be travelled in 4!/2!*2! ways. Number of ways to travel from the south-east corner of square P to B is 6!/4!*2!.
B to the north-east corner of D can be travelled in 6!/5! ways. From there to C can be travelled in 2 ways. There is 1 other way of travelling from B to C (along the perimeter of the field).
The total number of ways is, therefore, (4!/2!*2!) * (6!/4!*2!) * (6!/5! * 2 + 1)
= 6*15*13 = 1170

The nth term is a + (n-1)d
1000 = 1 + (n-1)d
So, (n-1)d = 999
999 = 3^3 * 37
So, the number of factors is 4*2 = 8
Since there should be at least 3 terms in the series, d cannot be 999.
So, the number of possibilities is 7

If the first task is assigned to either person 3 or person 4, the second task can be assigned in only 1 way. If the first task is assigned to either person 5 or person 6, the second task can be assigned in 2 ways. Therefore, the number of ways in which the first two tasks can be assigned is 2*1 + 2*2 = 6.
The other 4 tasks can be assigned to 4 people in 4! ways.
The total number of ways of assigning the 6 tasks is, therefore, 6*4! = 144.

Number of games in which both the players are girls = $$^GC_2$$ where G is the number of girls
$$^GC_2 = 45$$
$$^{10}C_2 = 45$$
So, G = 10
Similarly, number of games in which both the players are boys = $$^BC_2$$, where B is the number of boys
$$^BC_2 = 190$$
$$^{20}C_2 = 190$$
So, B = 20
So, number of games in which one player is a boy and the other player is a girl is 20*10 = 200

When the odd numbers occupy places 1 and 3, only 2 or 4 can be in the 5th place. Odd numbers can occupy places 1 and 3 in 3C2*2! = 6 ways. When 2 is at the 5th place, the other odd number and 4 can be arranged in the remaining places in 2 ways. So, 2 occurs at the end 6*2 = 12 times. Similarly, 4 occurs 12 times.

If odd numbers occupy places 1 and 5, then 2 or 4 should come in the 3rd place. The other two numbers can then be arranged in 2 ways in the remaining blanks. So, if 1 is in the first place and 5 is in the 5th place, the other numbers can be arranged in 2*2 = 4 ways. Similar for 1 and 3; 5 and 1; 3 and 1; 5 and 3; 3 and 5. So, 5 occurs 8 times, 1 8 times and 3 8 times. Similar is the case when odd numbers are placed in 3rd and 5th places.

On the whole, 4 occurs 12 times, 2 occurs 12 times, 5, 3 and 1 each occur 16 times. The total is, therefore, 48+24+80+48+16 = 216

Consider there are 6 people numbered 1-3 englishmen and 3-6 frenchmen, let 3 know both english and french.
First call would be between 1-3 then 2-3 such that 3 know secret of all 3 englishmen.
Let 3 call 4 .
Similarly there would be call between 4-5 then 4-6 such that 4 know secret of all 3 frenchmen.
Now 3 would call 4 . Such that 3 and 4 would know secret of all 6 members.
Now to let this know to 1,2,5,6 more 4 calls would be required.

Hence, minimum calls required would be 9.

Total number of pairs is $$^{N}C_2$$. Number of pairs standing next to each other = N. Therefore, number of pairs in question = $$^{N}C_2$$ - N

= 28/2 = 14.

If N = 7,

$$^{7}C_2$$ - 7 = 21 - 7 = 14.

N = 7.

The person has to take 3 steps north and 4 steps west, in whatever way he travels.

Total steps = 7, 3 north and 4 west.

Number of ways = 7!/(4!3!) = 35

The number of ways of selecting a colour for the first stripe is 4. The number of ways of selecting a colour for the second stripe is 3. Similarly, the number of ways of selecting colours for the third, fourth, fifth and sixth stripes are 3, 3, 3 and 3 respectively.
The total number of ways of selecting the colours is, therefore, 4*3*3*3*3*3 = 12*81.

 x, y and z in the hundred's, ten's and unit's place. So y should start from 2

If y=2 , possible values of x=1 and z = 0,1 .So 2 cases 120,121.

Also if y=3 , possible values of x=1,2 and z=0,1,2.

Here 6 three digit nos. possible .

Similarly for next cases would be 3*4=12,4*5=20,5*6=30,.....,8*9=72 . Adding all we get 240 cases. 

If there is only 1 green ball, it can be done in 6 ways
If there are 2 green balls, it can be done in 5 ways.
.
.
.
If there are 6 green balls, it can be done in 1 way.
So, the total number of possibilities is 6*7/2 = 21

Take any 12 points.

The maximum number of edges which can be drawn through these 12 points are $$^{12}C_2$$ = 66

The minimum number of edges which can be drawn through these 12 points are 12-1 = 11 as the resulting figure need not be closed. It might be open.

If there are 'm' non-parallel lines, then the maximum number of regions into which the plane is divided is given by

[m(m+1)/2]+1

In this case, 'm' = 10

So, the number of regions into which the plane is divided is (10*11/2) + 1 = 56

First a black square can be selected in 32 ways. Out of remaining rows and columns, 24 white squares remain. 1 white square can them be chosen in 24 ways. So total no. of ways of selection is 32*24 = 768.

According to given condiitons, number of 1 digit nos. = 2, number of 2 digit nos. = 2*3 , number of 3 digit nos. = $$2*3^2$$ , number of 4 digit nos. = $$2*3^3$$ , number of 5 digit numbers. = $$2*3^4$$,  Number of 6 digit nos. = $$2 *3^5$$.

Total summation 2*(1+3+9+27+81+243) = 728 . 

For any selection of three numbers, there is only one way in which they can be arranged in ascending order.

So, the answer is $$^{10}C_3 = 120$$

The number of ways in which this can be done is 11*10*9*8 = 7920

If there are 3 symmetric letters, it can be formed in 11*10*9 ways

If there are 2 symmetric letters, it can be formed in 11C2 * 15C1 * 3! ways

If there is only 1 symmetric letter, the password can be formed in 15C2*11C1*3! ways

Total = 990+330*15+630*11 = 12870 ways

The distinct paths are:

A -> D -> C -> F
A -> D -> E -> F
A -> D -> C -> E -> F
A -> B -> D -> C -> F
A -> B -> D -> E -> F
A -> B -> D -> C -> E -> F
A -> B -> C -> F
A -> B -> E -> F
A -> B -> F
A -> B -> C -> E -> F

So, the total number of distinct paths is 10

The three white flags can be arranged in the following two ways:

__ W __ W __ W or W __ W __ W __

In the blanks, the 2 blue and one red flag can be arranged in 3 ways.
So, the total number of arrangements is 2*3 = 6

Consider cases : 1) Last digit is 9: No. of ways in which the first 3 digits can be guessed is 2. No. of ways in which next 3 digits can be guessed is 9*9*9. So in total the number of ways of guessing = 2*9*9*9 = 1458.

2) Last digit is not 9: the number 9 can occupy any of the given position 4, 5, or 6, and there shall be an odd number at position 7.

So in total, the number of guesses = 2*3*(9*9*4) = 1944+1458 = 3402

The possible roads are:

A -> B Let this be x
A -> C Let this be y
B -> C Let this be z

From the information given, x + yz = 33 -> (1)
z + xy = 23 -> (2)

From the options, if y = 10, x =2 and z = 3 from (2), but it doesn't satisfy (1)
If y = 5, x = 4 and z = 3 from (2) but they don't satisfy (1)
A possible set of numbers for (x,y,z) are (3,6,5)
Number of roads from A -> C = 6

At least one candidate and at most n candidates among 2n+1 candidates => 

^{2n+1}C_1Total number of ways alteast one candidate can be selected = Total number of ways of selecting a candidate - total number of ways of selecting no candidates

$$\Rightarrow ^{2n+1}C_1+^{2n+1}C_2+^{2n+2}C_3+....+^{2n+1}C_{n-1}+^{2n+1}C_n = 63 $$

We know that $$^{2n+1} C_0$$ and $$^{2n+1} C_{2n+1}$$ are equal to 1.

By Binomial Expansion 

$$ ^{2n+1}C_0+^{2n+1}C_1$$+... + $$^{2n+1}C_{2n}+^{2n+1}C_{2n+1} = (1+1)^{2n+1} $$ ---- Eq 1

Also $$^{2n+1}C_1$$ = $$^{2n+1}C_{2n}$$ by symmetry

and $$^{2n+1}C_2$$ = $$^{2n+1}C_{2n-1}$$ and so on

So $$\Rightarrow ^{2n+1}C_{n+1}+^{2n+1}C_{n+2}+....+^{2n+1}C_{2n-1}+^{2n+1}C_{2n} = 63 $$

Therefore, on substituting these values in Eq 1 we get

1 + 63 + 63 +1 = $$2^{2n+1}$$

$$2^{2n+1}$$ = 128

2n+1 = 7

Therefore, n=3

As at most n students can be selected, the correct answer is 3.

Chariman, Vice-Chairman and the director can be made as a group such that Chairman sits between the Vice-Chairman and the director. This group can be formed in 2 ways.

Each of the remaining 7 directors and the group can be arranged in 7! ways.

=> Total number of ways = 2 * 7!.

Selecting 3 girls from 5 girls can be done in $$^5C_3$$ ways => 10 ways

Each of the boys may or may not be selected => 2 * 2 * 2 * 2 = 16 ways

=> 16 * 10 = 160 ways

For a number to be even, its unit digit should be 0,2,4,6,8
Case 1: One of the digit is 5
Hence according to question, 5 can't come in middle and at unit's place, so numbers will be 570,572,574,576,578.
Case 2: No digit is 5
Hence the hundreds place can be filled in 8 ways (except 0,5) and tens place can be filled in 9 ways (except 5).
Number of ways = 8 * 9 * 5 = 360
Hence total number of ways = 360 + 5 = 365

The expected money got by the player = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 21/6 = Rs 3.5

So, the player has to pay 3.5 - 1 = Rs 2.5 to get a profit of Re 1 in the long run.

Total ways when all 6 boxes have only black balls = 1
Total ways when 5 boxes have black balls = 2
Total ways when 4 boxes have black balls = 3
Total ways when 3 boxes have black balls = 4
Total ways when 2 boxes have black balls = 5
Total ways when only 1 box has black ball = 6
So total ways of putting a black ball such that all of them come consecutively = (1+2+3+4+5+6) = 21