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(41) = 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
 8C4
 9C4
 (12C4) – 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(51) = 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 x^{n} in (1x)^{(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 nonnegative 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 nonnegative 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 nonnegative 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 nonnegative 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 nonnegative 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)