Loading....
Coupon Accepted Successfully!

 

Division and Distribution of Objects

(With fixed number of objects in each group)

(i)Into groups of unequal size (different number of objects in each group)

  • Number of ways, in which n distinct objects can be divided into r unequal groups containing a1, a2, a3, ......, ar  things (a1 ≠ aj)

 = nCa1n-a1Ca2n-a1-a2Car. = n!/a1!a2!a3!...ar!

 Here a1 + a2 + a3 + ...... + ar = n.

  • Number of ways in which n distinct objects can be distributed among r persons such that some person get a1 objects, another person get a2 objects ......... and similarly someone gets ar objects = n!r! /a1!a2!a3!...ar!. 

Explanation: Let us divide the task into two parts. In the first part, we divide the objects into groups. In the second part, these r groups can be assigned to r persons in r! ways.

 

(ii) Into groups of equal size (each group containing same number of objects)

  • Number of ways in which m × n distinct objects can be divided equally into n groups (unmarked) = (mn)!/(m!)n n!.
  • Number of ways in which m × n different object can be distributed equally among n persons (or numbered groups)

= (number of ways of dividing) × (number of groups)! = (mn)!n!/(m!)n n! = (mn)!/(m!)n .

Some Usefull Tips

(i) Number of combinations of dissimilar things taken r at a time when p particular things always occur = n-pCr-p. 

Explanation: Here, actually we are making a selection of (r - p) things out of (n - p) things which can be done in n-pCr-p ways.


(ii) 
Number of permutations of n dissimilar things taken r at a time when p particular things always occur = n-pCr-pr!

Explanation: Here, number of combinations is the same as above but every combinations of r things can be permutated in r! ways and that's why total number of permutations = n-pCr-pr!


(iii) 
Number of combinations of n dissimilar things taken r at a time when p particular things never occur = n-pCr-p.

Explanation: Here, actual selection of r things is being made out of (n - p) things and that's why total number of selections = n-pCr-p.


(iv) 
Number of permutations of n dissimilar things taken r at a time when p particular things never occur = n-pCr-pr!.

Explanation: Here, number of combinations is the same as above but every selection made of r things can be permutated in r! ways and therefore then total no. of permutations = n-pCr-pr!.


(v) Gap Method:
 If there are m men and n women (m > n) and they have to sit in a row in such a way that no two women sit together then total no. of such arrangements = m+1Cn. m!

Explanation: If we denote men by m and women by w then there are exactly (m + 1) places in which women can be placed such that no two women will be together. This can be done in m+1Cn ways. Moreover, m men can be arranged among themselves in m! ways. Therefore, total number of arrangements : m+1Cn.m!

w m w m w .................. m w m w


(vi) String method:
 Many a times, one may encounter a problem of arranging n number of persons in a row such that m of them is always together. For this we tie all these m persons with a string i.e. we treat them as one, then we have (n - m + 1). Therefore total number of arrangements = (n - m + 1)!.m!.

Never together: If question is to find the number of arrangements such that m out of n are never all together then such total number of arrangements.

Total possible arrangements of a n persons without any restrictions - Total arrangements when m out of them are always together.

Solved Examples

Example

A letter lock consists of three rings each marked with 10 different letters. In how many ways is it possible to make an unsuccessful attempt to open the block?

Solution

Two rings may have same letter at a time but same ring cannot have two letters at time, therefore, we must proceed ring wise.
Each of the three rings can have any one of the 10 different letters in 10 ways.
Therefore Total number of attempts = 10 × 10 × 10 = 1000.
But out of these 1000 attempts only one attempt is successful.
Therefore Required number of unsuccessful attempts
=      1000 - 1 = 999.


 

Example

Find the total number of signals that can be made by five flags of different colour when any number of them may be used in any signal.

Solution

Case I :     When only one flag is used.
                No. of signals made = 5P1 = 5.
Case II :    When only two flag is used.
                Number of signals made = 5P2 = 5.4 = 20.
Case III :   When only three flags are used.
                Number of signals is made = 5P3 = 5.4.3 = 60.
Case IV :   When only four flags are used.
                Number of signals made = 5P4 = 5.4.3.2 = 120.
Case V :     When five flags are used.
                Number of signals made = 5P5 = 5! = 120.
Hence, required number = 5 + 20 + 60 + 120 + 120 = 325. 



 

Example

Prove that if each of the m points in one straight line be joined to each of the n points on the other straight line, the excluding the points on the given two lines. Number of points of intersection of these lines is 1/4 mn
(m-1(n-1).

Solution

To get one point of intersection we need two points on the first line and two points on the second line. These can be selected out of n-points in nC2 ways and for m points in mC2 ways.
Therefore Required number = mC2 × nC2
    = (m(m-1))/2! . (n(n-1))/2!
    = 1/4 m n (m - 1)(n - 1) 


 

Example

There are ten points in a plane. Of these ten points four points are in a straight line and with the exception of these four points, no other three points are in the same straight line. Find
(i) the number of straight lines formed.
(ii) the number of triangles formed.
(iii) the number of quadrilaterals formed by joining these ten points.

Solution

(i) For straight line, we need 2 points
 
 

No. of point selected out of 4 collinear points No. of points selected out of remaining 6 points No. of straight line formed
0
1
2
2
1
0
4C0 × 6C2 = 15
4C1 × 6C1 = 24
1
(In last case only one straight line is formed)
 Therefore Required number = 15 + 24 + 1 = 40
 
(ii)    For triangle, we need 3 points
No. of point selected out of 4 collinear points No. of points selected out of remaining 6 points No. of triangles formed
0
1
2
3
3
2
1
0
4C0 × 6C3 = 20
4C1 × 6C2 = 60
4C2 × 6C1 = 36
0
(In last case number of triangles formed is 0)
Therefore Required number = 20 + 60 + 36 + 0 = 116
 
(iii) For a quadrilateral, we need 4 points
No. of point selected out of 4 collinear points No. of points selected out of remaining 6 points No. of quadrilateral formed
0
1
2
3
4
4
3
2
1
0
4C0 × 6C4 = 15
4C1 × 6C3 = 80
4C2 × 6C2 = 90
0
0

(In these cases number quadrilateral is formed
Therefore Required number = 15 + 80 + 90 = 185.

 

Example

A student is allowed to select at most n-blocks from a collection of (2n + 1) books. If the total number of ways in which he can select a book is 63, find the value of n.

Solution

Since the student is allowed to select at the most n-books out of (2n + 1) books, therefore, he can choose, one book, two books or at the most n books. The number of ways of selecting at least one books are
        2n+1C1 + 2n+1C2 + ......... 2n+1Cn = 63 = S (Say)
Again, we know that
        2n+1C0 + 2n+1C1 + ......... 2n+1Cn + 2n+1C2n+1 = 22n+1
Now  2n+1C0 = 2n+1C2n+1 = 1
        2n+1C1 = 2n+1C2n etc.........
Hence, we have
        1 + 1 + 2S = 22n+1
or     2 + 2 . 63 = 22n+1
or     128 = 22n+1 or 27 = 22n+1
=>     2n+1 = 7
or     2n = 6
        n = 3

 

 

Example

A family consists of a grandfather, 6 sons and daughters and 5 grand children. They are to be seated in a row for dinner. The grand children wish to occupy the two seats at each end and the grandfather refuses to have a grandchild on either side of him. In how many ways can the seating arrangements be made for the dinner?

Solution

There are 6 adults, 4 grand children and 1 grand father.
Let us mark the seat for 11 persons from 1 to 11.
     S      S      S                      S      S      S
     1      2      3 .................9      10     11
Seats number 1, 2 and 10, 11 at the ends are to be occupied by 4 grand children and it can be done in 4P4 = 4! = 24 ways.
Now, we will seat the grandfather who cannot occupy seat number 3 or seat number 9 because he does not want to have a child by his side. Hence, he has to choose any of five seats 4, 5, 6, 7, 8 i.e. can seat himself in 5 ways.

 

Example

In an examination, the maximum marks for each of the three papers are 50 each. Maximum marks for the fourth paper is 100. Find the number of ways in which the candidate can score 60% marks in the aggregate.

Solution

Aggregate of marks 50 × 3 + 100 = 250
        Therefore 60% of the aggregate = 3/5 × 250 = 150
        Now the number of ways of getting 150 marks in the aggregate
        = coefficient of x150 in (x0 + x1 +.........+ x50)3 (x0 + x1 + x2 +......+ x100)
        = coeff. of x150 in ((1 - x51)/(1 -x))3 ((1 - x101)/(1 - x))
        = coeff. of x150 in (1-x51)3(1-x101)(1-x)-4
        = coeff. of x150 in (1-3x51 + 3.x102 - x153)(1-x101)(1-x)-4
            coeff. of x150 i
              
        = 1.(151.152.153)/6 - 3(100.101.102)/6 - 1(50.51.52)/6 + 3(49.50.51)/6
        = 151.76.51 - 100.101.51 - 50. 17.26 + 49.25.51
        = 57 (151.76 - 100 - 101) + 17 (49.25 - 50.26)
        = 51 (11476 - 10100) + 17 (3675 - 1300)
        = 51.1376 + 17.2375 = 70176 + 40375
        = 110551.

 

Example

Find the number of 4 lettered words that can be formed form the letters of the word PROPORTION?

Solution

Step 1              Letter                       Freq.
                          P                                2
                          R                                2
                          O                                3
                          T                                 1
                           I                                 1
                          N                                1
                        --------------------------------------------
                                                        10 (always check total)
Step 2
(i)     here we cannot have all the 4 letters alike, so
(ii)    3 alike               1 diff.                Selection                    Remark
        Say X                    Y              1C1 × 5C1 = 5              1C1 → 3O's
                                                   5C1 → one form {P, R, T, I, N}
Arrangement of these 3X's and 1 Y =  
                                                    = 4!/3!  = 4
Hence total such words = 5 × 4 = 20.
 
(iii)    2 same              2 diff.                                Selection
        2(X)                   (Y, Z)                                3C1 × 5C2 = 12
        Arrangement = (p+q+r)/p!q!r!  = (2+1+1)2!1!1! = 12
        Hence total words = 12 × 30 = 360
(iv)   2 same              2 another same           Selection
        (2X)                  (2Y)                          3C2 = 3
        Arrangements = (2+2)!/2!2!  = 6
        Total words = 3 × 6 = 18
(v)    1 same, 3 diff º 4diff. Selection 6C4 = 6C2 = 15
        Arrangement = (1+1+1+1)!/1!1!1!1!  = 4! = 24
        Total words = 15 × 24 = 360
(vi)   1 same, 3 another same º case ii (already considered)
        Hence grand total of desired words
                = 20 + 360 + 18 + 360 - 758.

 

 

Example

You are given the responsibility of organizing a fresher's welcome party at IIT-Delhi. Total 496 students will join IIT-D. Six restaurants A, B, C, D, E and F are booked. Capacity of each is 120, 146, 46, 72, 72, 80 persons at a time respectively. You have to group them in such a way that each group has least possible number of students. How would you do so?

Solution

But:  496/6 = 82.6 > 46
Therefore C has least capacity of 46 and is not capable to accommodate 82 students. Hence let us group for it first i.e. 46 students get accommodated.
Students left                       Restaurant left
         450                           A B D E and F
But 450/5 = 90 > 72
Next D and E each require smallest group 72, 72
             D → 72                              E → 72
           Students left                       Restaurant left
               306                               A B and F
Now F is the least left for grouping, so assigning 80 to F.
         Students left                       Restaurant left
               226                               A and B
Now Therefore 226/2 = 113 < 120 and each of A and B is capable of accommodating to this. Hence,
        A → 113
        B → 113
But not
        A → 120
        B → 106
As in the case A does not have minimum possible number of students.
Hence, we grouped 496 students into 46, 72, 72, 80, 113 and 113 each.
No. of ways of doing so = 496! / (46!(72!)2 .2!80!(113!)2.2!)
Remember: Equal division minimizes the number of students in each group.

 
 

Example

How many different numbers can be formed with the digits 1, 3, 5, 7, 9 when taken all at a time and what is their sum?

Solution

The total number of numbers = |5 = 120. Suppose we have 9 in the unit's place. We will have |4 = 24 such numbers. The number of numbers in which we have 1, 3, 5 or 7 in the unit's place is |4 = 24.
Hence, the sum of the digits in the unit's place in all the 120 numbers
        = 24 (1 + 3 + 5 + 7 + 9)
        = 600.
The number of numbers when we have any one of the given digits in ten's place is also |4 = 24 in each case. Hence, the sum of the digits in the ten's place = 24 (1 + 3 + 5 + 7 + 9) tens
        = 60 tens = 600 × 10.
Proceeding similarly, the required sum
= 600 units + 600 tens + 600 hundreds + 600 thousands + 60 ten thousands.
= 60 (1 + 10 + 100 + 1000 + 10000) = 600 × 11111
= 6666600.

 

 

Example

Find the values of r for which the number fo combinations of n things taken are at a time is greatest.

Solution

Since         nCr = (n(n-1)(n-2)....(n-r+2)(n-r+1))/(1.2.3...(r-1)r)
and           nCr-1 = (n(n-1)(n-2)....(n-r+2))/(1.2.3...(r-1))
Therefore    nCr = nCr-1 × (2-r+1)/r = nCr-1(((n+1)/r)-1).
The multiplying factor (n-r+1)/r may be written as (((n+1)/r) - 1), which shows that it decreases as r increases. Hence as r resumes the values 1, 2, 3,...... in succession, nCr is continually increased until (((n+1)/r) - 1) becomes equal to 1 or less than 1.
Now (((n+1)/r) - 1) > 1
so long as (n+1)/r) - 2
that is (n+1)/r) - r .
We have to choose the greatest value of r consistent with this inequality.

​(1) Let n be even, and equal to 2m, then
(n+1)/2 = (2m+1)/2 = m+(1/2)  ;
and for all values r up to m (inclusive of m) this is greater than r.
Hence by putting r = m = n/2 , we find that the greatest number of combination isnCn/2.
 
(2) Let n be odd, and equal to 2m + 1; then
(n+1)/2 = (2m+1)/2 = m+1 ;
and for all values of r up to m (inclusive of m) this is greater than r, but when r = m + 1 the multiplying factor becomes equal to 1, and nCm+1 = nCm; that is
n+1nC/ 2 = n+1nC/ 2 and
Therefore the number of combinations is greatest when the things are taken (n+1)/2 or (n-1)/2 at a time, the result being the same in the two cases.





Test Your Skills Now!
Take a Quiz now
Reviewer Name