# Permutation & Combination concepts by Gaurav Sharma - Part (4/5)

• In how many ways can 3 boys and 15 girls sit together in a row such that between any 2 boys at least 2 girls can sit?

Firstly, 3 boys can be arranged in 3! Ways

(X ) [B] (Y) [B] (Z) [B] (W)

After arranging the boys 4 gaps are created. Let in these gaps x, y, z and w girls sit.

Now y, z ≥ 2 and x, w ≥ 0

x + y + z + w = 15

Let y = y1 + 2 and z = z1 + 2 (where y1, z1 ≥ 0)

x + y1 + z1 + w = 11

F

Number of solutions = (11 + 4 – 1)C(4-1) = 14C3

Now the girls can be arranged in 15! Ways

So total ways are 14C3 x 3! X 15!

Between 2 junctions A and B there are 12 intermediate stations. The number of ways in which a train can be made to stop at 4 of these stations, so that no two halting stations are consecutive is

1. 8C4
2. 9C4
3. (12C4) – 4
4. None of these

(A) [X1] (1) [X2] (2) [X3] (3) [X4] (4) [X5] (B )

Here x1 + x2 + x3 + x4 + x5 = 8 -- > (1)

x1, x2, x3, x4, x5 is the number of intermediate stations.

1, 2, 3, 4 are the stations.

Here x1 ≥ 0; x5 ≥ 0; x2, x3, x4 ≥ 1

The total number of ways is the number of solutions of the above equation.

Let v = x2 – 1, u = x3 – 1, w = x4 – 1

We get, x1 + v + u + w + x5 = 5 (from equation 1)

Where v, u, w ≥ 0

The number of solutions = (5 + 5 – 1)C(5-1) = 9C4

 If n objects are arranged in a row, the number of ways of selecting r items such that no two items are consecutive (n – r + 1) C r
 Coefficient of xn in (1-x)(-r) is (n + r – 1)C(r – 1)

In how many ways can a sum of 10 be obtained by three persons, each throwing a single dice once?

Method 1:

A + B + C = 10 where 1 ≤ A, B, C ≤ 6

Now let A = a + 1, B = b + 1 and C = c + 1 where 0 ≤ a, b, c ≤ 5

So a + 1 + b + 1 + c + 1 = 10

a + b + c = 7

Number of non-negative solutions for the equation = (7 + 3 – 1)C(3 – 1) = 9C2

Now this will include case when a > 5, so put a = 6 + a1

6 + a1 + b + c = 7

a1 + b + c = 1

Number of non-negative solutions for the equation = ( 1 + 3 – 1) C (3 – 1) = 3

Similarly for b and c also there will be 3 cases each

So total solutions = 9C2 – 9 = 36 – 9 = 27

Method 2 :

A + B + C = 10

Put A = 6 – a, B = 6 – b, C = 6 – c

We get 6 – a + 6 – b + 6 – c = 10

a + b + c = 8

Total 10C2 = 45

Subtract cases where a ≥ 6, put a = 6 + a1

a1 + b + c = 2

Total 6 cases

Similarly for b ≥ 6 and c ≥ 6, 6 cases each.

So total cases = 45 – ( 6 x 3 ) = 45 – 18 = 27

In how many ways can a sum of 15 be obtained by three persons, each throwing a single dice once?

Let A, B and C be the three persons.

Now A + B + C = 15

Also, A ≤ 6, B ≤ 6 and C ≤ 6

So, put A = 6 – a, B = 6 – b and C = 6 – c

6 – a + 6 – b + 6 – c = 15

a + b + c = 3

Hence number of non-negative solutions of a + b + c = 3 are

(3 + 3 – 1) C (3 – 1) = 5C2 = 10

In how many ways can a sum of 11 be obtained by three persons, each throwing a single dice once?

Let A, B and C are the three persons

Now, A + B + C = 11

Also, A ≤ 6, B ≤ 6 and C ≤ 6

So, put A = 6 – a, B = 6 – b and C = 6 – c

6 – a + 6 – b + 6 – c = 11

a + b + c = 7

Hence number of non-negative solutions of a + b + c = 7 are

(7 + 3 – 1) C (3 – 1) = 9C2 = 36

But if (a =0, b=1,c=6) then C = 6 – c = 0

Which is not possible and there are 9 such cases

So number of ways = 36 – 9 = 27

Find the number of ways in which a score of 21 can be made from a throw by 4 persons, each throwing a single dice.

We have A + B + C + D = 21

So, put A = 6 – a, B = 6 – b, C = 6 – c and D = 6 - d

6 – a + 6 – b + 6 – c + 6 - d = 21

a + b + c + d = 3

Hence number of non-negative solutions of a + b + c + d = 3 are

(3 + 4 – 1) C (4 – 1) = 6C3 = 20

Find the number of ways in which we can get a sum greater than 17 by throwing six distinct dice.

Let a, b, c, d, e and f be the numbers that appear on the six dice.

Here 1 ≤ a,b,c,d,e,f ≤ 6

Total number of solutions = 6^6

To find solutions for a + b + c + d + e + f > 17

Number of ways to get the sum less than or equal to 17 = 17C11 – 6 x 11C5

So number of ways to get the sum greater than 17 is 6^6 – (17C11 – 6 x 11C5)

• In how many ways can a sum of 11 be obtained by three persons, each throwing a single dice once?
Hi,Could you please explain the above problem @gaurav_sharma

• @cat789

It is explained in great detail here - https://www.mbatious.com/topic/887/solving-combinatorics-problems-using-stars-bars-concept/

Please have a look and if the doubt persist, let us know. Will help in our best capacity. Happy Learning!

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