PERMUTATION AND COMBINATION 1

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

The fundamental principle of counting
Sol.
1. There are 4 letter boxes in a post office. In how many ways
can a man post 8 distinct letters ?
(a) 4 × 8 (b) 84
(c) 48 (d) P (8, 4)
Ans. (c) 1st box can be filled in 4 ways (any colour)

Sol. Each letter has 4 ways to get posted in the post office. All box from 2nd onwords can be filled by 3 ways leaving
the colour in the previous box.
 Total No. of ways  48
 total no. of flag  4  3  3  3  3  3
2. In an examination there are three multiple choice questions
and each question has 4 choices out of which only one is  12  81
correct. If all the questions are compulsory, then number of 5. 4 buses run between Bhopal and Gwalior. If a man goes
ways in which a student can fail to get all answers correct, from Gwalior to Bhopal by a bus and comes back to Gwalior
is by another bus, then the total possible ways are
(a) 11 (b) 12 (a) 12 (b) 16
(c) 27 (d) 63 (c) 4 (d) 8
Ans. (d) Ans. (a)
Sol.  Total possible ways to answer  4  4  4  64
Sol. Bhopal to Gwalior  4buses
 Only 1 way is there to answer all questions correctly For that man:
 Required no. of ways  64  1  63 Gwalior to Bhopal : 4buses
3. Every one of the 5 available lamps can be switched on to
Then Bhopal to Gwalior : 3buses
illuminate certain Hall. The total number of ways in which
the hall can be illuminated, is : So,Total ways  4  3  12 ways
(a) 32 (b) 31
Arrangement & selection for different objects
(c) 5 (d) 5!
Ans. (b) 6. How many numbers lying between 500 and 600 can be
formed with the help of the digits 1, 2, 3, 4, 5, 6 when the
Sol. Required no. of ways  Total no. of positions of 5 switches
digits are not to be repeated
-[the no. of case when no lamp is
(a) 20 (b) 40
switched ON]
(c) 60 (d) 80
 (2  2  2  2  2)  (1) Ans. (a)
 32  1  31 Sol. Digit at 100 th place can be filled in 1 way ( As only 5 can be
4. A new flag is to be designed with six vertical strips using there)
some or all of the colours yellow, green, blue and red. Then,
Digit at 10 th and units can be filled in 5  4 ways
the number of ways this can be done such that no two
adjacent strips have the same colour is So, Answer = 1 ×5 × 4 = 20 ways
(a) 12 × 81 (b) 16 × 192 7. The number of 3 digit odd numbers, that can be formed by
(c) 20 × 125 (d) 24 × 216 using the digits 1, 2, 3, 4, 5, 6 when the repetition is allowed,
is
Ans. (a)
(a) 60 (b) 108
(c) 36 (d) 30
 2 PERMUTATION AND COMBINATION
Ans.(b) Ans. (c)
Sol. Unit place can be filled in 3 ways Sol. No. of all 3 digit numbers having no digit as 5 are

10 th place and 100 th place can be filled in 6  6 ways

So, Answer  3  6  6  108 ways

8. The number of all four digit numbers is equal to
(a) 9999 (b) 9000
(c) 104
(d) none of these
 648
Ans.(b)
12. All possible three digits even numbers which can be formed
with the condition that if 5 is one of the digit, then 7 is the
Sol. next digit is :
(a) 5 (b) 325
(c) 345 (d) 365
Ans. (d)
 no. of 4 digit nos  9000
Sol. There will be two cases
9. The number of all four digits numbers with distinct digits is
Case - I : When 5 is not there
(a) 9×10×10×10 (b) 10P4
(c) 9 × 9P3 (d) none of these
Ans. (c)

Sol.

= 8 × 9 × 5 = 360
case- II :when 5 is there

 no. of 4 distinct 4 digit numbers  9 9 P3

10. The number of even numbers that can be formed by using
the digits 1, 2, 3, 4 and 5 taken all at a time (without repetition)
is
(a) 120 (b) 48
(c) 1250 (d) none of these 5
Ans. (b)
 Total numbers  360  5
 365
Sol.
13. How many of the 900 three digit numbers have at least one
even digit ?
(a) 775 (b) 875
(c) 450 (d) 750
 2  4! Ans. (a)

 total even numbers  48 Sol. There digit numbers which don’t have any even digit

11. The number of all three digit numbers having no digit as 5 is  5  5  5  125
(a) 252 (b) 225  Numbers having at last one even digit  900  125
(c) 648 (d) none of these  775
PERMUTATION AND COMBINATION 3

14. The number of six digit numbers that can be formed from the removed we get the numbers  8! 7!
digits 1, 2, 3, 4, 5, 6 & 7 so that digits do not repeat and the
terminal digits are even is :  Total 8 digit numbers  8! 4(8! 7!)
(a) 144 (b) 72  5(8!)  4(7!)
(c) 288 (d) 720
 7!(5  8  4)
Ans. (d)
Sol. (Since terminal digits are fixed i.e, even. so 5 numbers at  7! 36
10th place can be permuted in 5 ways) 16. 8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4,
Required Numbers 4. The number of such numbers in which the odd digits do
not occupy odd places, is :
(a) 160 (b) 120
(c) 60 (d) 48
Ans. (b)

Sol. Odd digits 1,1,3 to be arranged at four even

places.
 6  120  720
15. An eight digit number divisible by 9 is to be formed using This can be done in 4 C3  3!/ 2!  12
digits from 0 to 9 without repeating the digits. The number
of ways in which this can be done is: Next, the even numbers  2, 2, 2, 4, 4  go to

(a) 72  7  (b) 18  7  the remaining 5 places in 5!/ (3!.2!) ways  10 ways

So: required number of ways  12  10  120 ways

(c) 40  7  (d) 36  7 
17. Two women and some men participated in a chess
Ans. (d) tournament in which every participant played two games
Sol. Sum of 0,1,2,3,4,5,6,7,8,9 is 45 with each of the other participants. If the number of games
that the men played between themselves exceeds the
 divisible by 9 number of games that the men played with the women by
66, then the number of men who participated in the
Let S  {0,1, 2,3, 4, 5, 6, 7,8, 9} tournament lies in the interval:
Now, (a) [8, 9] (b) [10, 12)
For 8 digit numbers we need to remove (c) (11, 13] (d) (14, 17)
two digits from ‘S’ such that the remaining digits Ans. (b)
must have sum divisible by 9 . Sol. Let the no. of men in the tournament =n
The no. of games that men played
 we can remove the pairs  0,9 or
between themselves  n C2  2
(1,8) or (2,7) or (3,6) or (4,5)
No. of games that men played with
If (0,9) are removed then no. of 8 digit
women  n C1 2 C1  2
numbers  8!
A.T.Q. n C2 ×2!= n C1×2 C1×2+66
If 1,8  are removed then no. of 8 digit

numbers  8! 7! n(n  1)  n  2.2  66

Similarly, when (2,7),(3,6) or (4,5) are n(n  1)  4n  66

n 2  5n  66  0
 4 PERMUTATION AND COMBINATION

(n  11)(n  6)  0  the number of selection of 3person in which 

- a couple is included

n  11  
18. On the occasion of Deepawali festival each student of a
class sends greeting cards to the others. If there are 20  6 C3  3 4 C1
students in the class, then the total number of greeting  20  3  4
cards exchanged by the students is
 20  12  8
20
(a) C2 (b) 2 .20 C 2
21. 5 Indian and 5 American couples meet at a party and shake
hands. If no wife shakes hands with her own husband and
(c) 2.20 P2 (d) None of these
no Indian wife shakes hands with a male, then the number
Ans. (b) of hand shakes that takes place in the party is :
Sol. Total number of exchanged greeting cards among (a) 95 (b) 110
(c) 135 (d) 150
' 20' students  2 20 C2
Ans. (c)
( 20 C2  selection of two students out of 20 students
Sol. No. of hand shakes  ( selection of 2 from 20) -
2  Number of exchanged greeting cards in between
( hand shakes by wife with own husband) -(hand shake by
these two students)
Indian wife with male)
So, Answer  2  C2
20
 20 C2  10  5  9  135
19. In a touring cricket team, there are 16 players in all including
22. A rack has 5 different pairs of shoes. The number of ways in
5 bowlers and 2 wicket-keepers. How many teams of 11
which 4 shoes can be chosen from it so that there will be no
players from these, can be chosen, so as to include three
complete pair is
bowlers and one wicket-keeper
(a) 1920 (b) 200
(a) 650 (b) 720
(c) 110 (d) 80
(c) 750 (d) 800
Ans. (d)
Ans. (b)
Sol. No.of selection  ( selection of 4 pairs from 5 pairs) 
Sol. Selection of 3 bowlers out of 5 bowlers 5 C3  10
(selection of 1 shoe from each selected pair)
Selection of 1 wicket-keeper out of 2  2 C1
5 C4  2  2  2  2  80
Selection of 7 batsman out of 9 players  C7  36
9
23. In how many ways can six boys and five girls stand in a row
Total number of selection  10  2  36 if all the girls are to stand together but the boys cannot all
stand together ?
 720ways
(a) 172,800 (b) 432,000
20. Three couples (husband and wife) decide to form a
(c) 86,400 (d) None of these
committee of three members. The number of different
committee that can be formed in which no couple finds a Ans. (b)
place is : Sol consider all girl as a single unit.
(a) 60 (b) 12 Now , B  B  B  B  B  B
(c) 27 (d) 8 X position are available for all girls together.
Ans. (d) Now, arrangements of boys =6!
Sol. Required Selection Position available for all girls together =5
=(Total number of selection of 3 persons out of 6 ) Arrangement of girls =5!
Required answer  6! 5  5!
PERMUTATION AND COMBINATION 5

 432000 (c) 56 (d) None of these

24. There are five different green dyes, four different blue dyes Ans. (c)
and three different red dyes. The total number of Sol. Required no. of times
combinations of dyes that can be chosen taking at least one
green and one blue dye is 8! 8.7.6
 8C3    56
(a) 3255 (b) 2
12 3!5! 3  2  1

(c) 3720 (d) none of these 27. In how many ways can two balls of the same colour be
selected out of 4 distinct black and 3 distinct white balls
Ans. (c)
(a) 5 (b) 6
Sol. At least 1 green dye can be chosen in
(c) 9 (d) 8
5
C1  C2  C3  C4  C5  2  1 ways
5 6 5 5 5
Ans. (c)
at least 1 blue dye can be chosen in Sol. Required solution = selection of two black balls + selection
of white balls
4
C1  4 C2  4 C3  4 C4  24  1 ways
 4 C2  3 C 2
Any no. of red dyes can be chosen in
 63  9
3
C0 3 C1  3 C2 3 C3  23 ways
28. If the letters of the word SACHIN are arranged in all
So , total no of required selection possible ways and these words are written out as in
dictionary, then the word SACHIN appears at serial number
     
 25  1  24  1  23  3720
(a) 602 (b) 603
25. Given 6 different toys of red colour, 5 different toys of blue (c) 600 (d) 601
colour and 4 different toys of green colour. Combination of
Ans. (d)
toys that can be chosen taking at least one red and one blue
toys are : Sol. A, C, H, I, N, S

(a) 31258 (b) 31248 No of words start with A=5!

(c) 31268 (d) None of these No of words start with C  5!
Ans. (b) No of words start with H  5!
Sol.  At least 1 red toy can be selected out of 6 different
No of words start with I=5!
toys in 6 C1  6 C2  6 C3 . 6 C6  64  1  63 ways No of words start with N  5!
at least 1 blue toy out of 5 different blue toys in Total words  5! 5! 5! 5! 5!  5(5!)  600
5
C1 5 C2 5 C5  32  1  31 ways Now add the rank of SACHIN so required rank of SACHIN
 600  1  601
 After selecting at least 1 green toy and 1 blue toy ,
selection of red toys (no restriction) 29. If the letters of the word ‘MOTHER’ are written in all possible
orders and these words are written out as in a dictionary,
 4 C0  4 C1  4 C2  4 C4  16 ways find the rank of the word ‘MOTHER’.
(a) 307 (b) 308
 Total no. of ways of selection
(c) 309 (d) 120
 63  31 16  31248 ways
Ans. (c)
26. A father with 8 children takes them 3 at a time to the
Zoological gardens, as often as he can without taking the Sol. In alphabetical order the letters of MOTHER are written as
same 3 children together more than once. The number of E,H,M,O,R,T
times he will go to the garden is
No. of words starting with E =120
(a) 336 (b) 112
No. of words starting with H =120
No. of words starting with ME =24
 6 PERMUTATION AND COMBINATION
No. of words starting with MH =24 Now, 49 th word NAAGI
No. of words starting with MOE =6
50th word NAAIG
No. of words starting with MOH =6
32. If all the letters of the word “QUEUE” are arranged in all
No. of words starting with MOR =6 possible manner as they are in a dictionary, then the rank of
No. of words starting with MOTE = 2 the word QUEUE is :
th th
No. of words starting with MOTHER = 1 (a) 15 (b) 16
th th
Rank = 309 (c) 17 (d) 18

30. The letters of the word RANDOM are written in all possible Ans. (c)
orders and these words are written out as in a dictionary Sol. E,E,Q,U,U
then the rank of the word RANDOM is
4!
(a) 614 (b) 615 Words starts with E   12
2!
(c) 613 (d) 616
Ans. (a) The Q E E U U 13th
Sol. A,D,M,N,O,R Q E U E U 14th
No, of words starting from (A,D,M,N,O)
Q E U U E 15th
 5  5  600
Q U E E U 16 th
No. of words starting from RAD

 36 Q U E U E 17 th

No, of words starting from RAM 33. The letters of word “RADHIKA” are permuted and arranged
in alphabetical order as in English dictionary. The number
 36 of words the appear before the word “RADHIKA” is :
No. of words starting from RANDM (a) 2193 (b) 2195
(c) 2119 (d) 2192
 11
Ans. (a)
No. of words starting from RANDOM
Sol. A, A, D, H, I, K, R
1
No. of words starting with ‘A’  6!  720
Total  614
31. If all permutations of the letters of the word AGAIN are 6!
No. of words starting with ‘D’=  360
arranged as in dictionary, then fiftieth word is 2!
(a) NAAGI (b) NAGAI
6!
(c) NAAIG (d) NAIAG No. of words starting with ‘H’   360
2!
Ans. (c)
Sol. A,A,G,I,N 6!
No. of words starting with ‘I’   360
2!
Now , starting with A  4!  24
6!
4! No. of words starting with ‘K’   360
starting with G   12 2!
2!
No. of words starting with ‘RAA’  4!  24
4!
Starting with I   12 No. of words starting with ‘RADA’  3!  6
2!
No. of words starting with ‘RADHA’  2!  2
24  12  12  48
PERMUTATION AND COMBINATION 7

No. of words starting with ‘RADHIA’  1

(n  1) n(n  1) n(n  1)(n  2)
   10
Next word is ‘RADHIKA’ 6 6
Total  720  360  360  360  360  24  6  2  1
 3n( n  1)  60  n 2  n  20  0
 2193
 (n  5)(n  4)  0,  n  5
Geometrical counting problems
37. There are 10 points in a plane, out of these 6 are collinear.
34. Number of rectangles in figure shown which are not squares If N is the number of traingles formed by joining these
is : points, then
(a) N > 190 (b) N  100
(c) 100 < N  140 (d) 140 < N  190
Ans. (b)

Sol. Number of required triangles 10 C3 6 C3

10  9  8 6  5  4
 
6 6
(a) 159 (b) 160  120  20
(c) 161 (d) None of these
 100
Ans. (b)
Hence option 2 is correct answer..
Sol. Required number of rectangles in the figure
38. Number of diagonals of a convex hexagon is
 Total possible rectangles - Total number of possible
(a) 3 (b) 6
squares
(c) 9 (d) 12
  7 C 2 5 C 2   [(4  6)  (3  5)  (2  4)  (1  3)]
Ans. (c)
 210  50 Sol. No. of sides , (n)  6 .Hence no. of diagonal
 160
 n C2  n
35. There are n points on a circle. The number of straight lines
formed by joining them is equal to  6 C2  6  9 (no. of diagonals)
(a) C2n
(b) P2
n
39. The interior angles of a regular polygon measure 150° each.
(c) C2 – 1
n
(d) none of these The number of diagonals of the polygon is
Ans. (a) (a) 35 (b) 44
Sol. Straight line can be drawn with two points (c) 54 (d) 78
 no. of straight lines  no. of ways of selection of 2 points Ans. (c)
from n
Sol. Interior angle  1500
 C2
n

 exterior angle  300

36. Let Tn be the number of all possible triangles formed by
joining vertices of an n-sided regular polygon. If 360
Tn + 1 – Tn = 10, then the value of n is  no. of sides   12
30
(a) 7 (b) 5
(c) 10 (d) 8  no. of diagonals  n C2  n 12 C2  12  54
Ans. (b) 40. Out of 10 points in a plane 6 are in a straight line. The

Sol. n+1
C3  n C3 =10
 8 PERMUTATION AND COMBINATION
number of triangles formed by joining these points are
And four S can be put in 8 places in 8 C4 ways.
(a) 100 (b) 150
The required number of ways
(c) 120 (d) None of these
Ans. (a)  7.5.3.8 C4  7.6 C4 8 C4

Sol. Total no. of triangles formed 10 C3 6 C2 44. The number of all possible different arrangements of the
word “BANANA” is
= 120 – 20 = 100
(a) 6 (b) 6  2  3
41. The number of straight lines that can be formed by joining
20 points no three of which are in the same straight line
6
except 4 of them which are in the same line (c) 2 3 (d) none of these
(a) 183 (b) 186
Ans. (c)
(c) 197 (d) 185
Ans. (d) Sol. A  3 times , N  2 , B  1 times.

Sol. w.r.t. the given condition 6

Required no. of straight lines  No. of arrangements 
23

 20 C2 4 C2  1 45. The total number of ways of arranging the letters AAAA

BBB CC D E F in a row such that letters C are separated from
 190  6  1  185 one another is
42. There are n distinct points on the circumference of a circle. (a) 2772000 (b) 1386000
The number of pentagons that can be formed with these
(c) 4158000 (d) none of these
points as vertices is equal to the number of possible
triangles. Then the value of n is Ans. (b)
(a) 7 (b) 8 10!
Sol. Total no, of ways  C2   1386000
11
(c) 15 (d) 30 4!3!
Ans. (b) 46. A library has ‘a’ copies of one book, ‘b’ copies of each of
Sol. No. of Pentagons  No. of triangles (Given) two books, ‘c’ copies of each of three books, and single
copy each of ‘d’ books. The total number of ways in which
 n C5 n C3 these books can be arranged in a row is

If n C x  n C y  x  y  n (a  b  c  d)! (a  2b  3c  d)!
(a) (b) a! b!  c!
2 3
a!b!c!
n=5+3=8

Arrangement & selection of like objects (a  2b  3c  d)!

(c) (d) none of these
a!b!c!
43. How many different words can be formed by jumbling the
Ans. (b)
letters in the word MISSISSIPPI in which no two S are
adjacent ?
(a  2b  3c  d )!
(a) 7.6C4 . 8C4 (b) 8. 6C4 . 7C4 Sol. Total no. of ways 
a !(b !)2 (c !)3
(c) 6.7. 8C4 (d) 6.8. 7C4
47. A question paper on mathematics consists of twelve
Ans. (a) questions divided into three parts. A, B and C, each
Sol. Leaving S, we have 7 letters M. I, I, I, P P, I. containing four questions. In how many ways can an
examinee answer five questions, selecting atleast one from
7 each part.
ways of arranging them  2 4  7.5.3
(a) 624 (b) 208
PERMUTATION AND COMBINATION 9

(c) 2304 (d) none of these Sol. M : 1 time

Ans. (a) I : 4 times
Sol. CaseI: 3,1,1 S : 4 times
3 P : 2 times
No. of ways  2  C 3  C 1  C1
4 4 4

Total no. of required selections

 192  (1  1)  (4  1)  (4  1)  (2  1)  1
Case II : 2, 2,1  2  5  5  3  1  149

3 4 Distribution of different objects

No. of ways  2  C 2  C 2  C1
4 4

51. The set S = {1, 2, 3,..., 12} is to be partitioned into

 432 three sets A, B, C of equal size. Thus, A B  C = S, A
B = B C = A  C = . The number of ways to partition
 total no. of ways  192  432  624 S is
48. Number of ways in which 4 boys and 2 girls (all are of (a) 12!/3!(4!)3 (b) 12!/3!(3!)4
different heights) can be arranged in a line so that boys as
well as girls among themselves are in decreasing order of (c) 12!/(4!)3 (d) 12!/(3!)4
height (from left to right), is : Ans. (c)
(a) 1 (b) 6! Sol. The number of ways of partition S is:
(c) 15 (d) None of these
12!
12
C 4 8 C 4 4 C4 
Ans. (c) (4!)3
Sol. Order is defined, so we are supposed to give a no. of
52. In an election three districts are to be canvassed by 2, 3 and
selection for the seating arrangements for the 4 boys 5 men respectively. If 10 men volunteer, the number of ways
and 2 girls out of 6 seats. they can be alloted to the different districts is :

So, required no. of ways 6 C4 2 C2  15 10! 10!

(a) 2! 3! 5! (b) 2! 5!
49. The total number of selections of atleast one fruit which can
be made from 3 bananas, 4 apples and 2 oranges is 10! 10!
(a) 39 (b) 315 (c) 2! 2 5! (d) 2! 2 3! 5!
   
(c) 512 (d) none of these
Ans. (a)
Ans.(d)
Sol. It’s the same way like distribution of 10 objects in 2 ,
Sol. Total selection such that at least one fruit is chosen
will be 3 and 5 groups

(3  1)(4  1)(2  1)  1 10!

 Number of ways 
2!3!5!
 4  5  3 1
53. The number of ways to give 16 different things to three
 59
persons A, B, C so that B gets 1 more than A and C gets 2
50. The total number of different combinations of one or more more than B, is :
letters which can be made from the letters of the word
‘MISSISSIPPI’ is 16!
(a) (b) 4!5!7!
(a) 150 (b) 148 4!5!7!

(c) 149 (d) None of these 16!

Ans. (c) (c) (d) 3!5!8!
3!5!8!
 10 PERMUTATION AND COMBINATION
Ans. (a)
12
Sol. w.r.t. the given data Sol. No. of ways  8 4  2
Let A gets x no. of objects
Note:( any friend can receive 8 or 4 . Hence multiplied by
Let B gets y no. of objects
Let C gets z no. of objects 2)

A.T.Q. x  y  1, z  y  2 Distribution of alike objects

x  y  z  16 56. The number of ways of distributing 8 identical balls in 3
distinct boxes, so that none of the boxes is empty, is
 y  1  y  y  2  16
(a) 5 (b) 21
3 y  15
(c) 38 (d) 8C3
y5 Ans. (b)

 x  4, z  7 Sol. i  Each box must contain at least one ball since no box
remains empty so we have the following cases
16!
Required no. of ways  Box Number of balls
4!.5!.7!
54. The number of ways in which 52 cards can be divided into 4
sets, three of them having 17 cards each and the fourth one
having just one card

52 ! 52 !
(a) (b)
(17!)3 (17!)3 3!
1 3!
51! 51!  Number of ways  3   3! 2
(c) (d) 2!
(17!)3 (17!)3 3!
 9  6  2  21
Ans. (b)
Sol. 17,17,17,1 (division in group)
As have case
1 52
No. of ways  3  C17  C17  C17  C1
35 18 1

ways and

52 have equal number of ways of arranging the balls different


 17)3  3 boxes.

55. The number of ways in which 12 balls can be divided ii  Let the number of balls in the boxes are x,y,z
between two friends, one receiving 8 and the other 4, is
respectively then x+y+z=8 and no
12! 12!2!
(a) (b) box is empty so each x,y,z  1
8!4! 8!4!
 I  m  n  3  8 where I=x-1,m=y-1,n=z-1
12!
(c) (d) none of these
8!4!2! i.e. (l  1)  ( m  1)  ( n  1)  8 are non negative integers
Ans. (b)
 Required number of ways  n  r 1 Cr
PERMUTATION AND COMBINATION 11

= 3+5-1 C5 = 7 C5 = 7 C2   u1  1   u 2  2    u 3  3   u 4  4   25

57. The total number of ways in which 11 identical apples can  u1  u 2  u 3  u 4  15

be distributed among 6 children is that at least one apple is
given to each child where u1  0, u 2  0, u 3  0, u 4  0
(a) 252 (b) 462 So, required distribution
(c) 42 (d) none of these
 15 41 C4 1
Ans. (a)

Sol. Here total ways will be like x i  1 and 18 C3

2
x1  x2  x3  x4  x5  x6  11 60. The total number of ways in which n number of identical
balls can be put in n numbered boxed (1, 2, 3, ......... n) such
th
Thus no. of ways  6 5 1 C5  252 that i box contains at least i number of balls, is :

(a) n2 (b) n 2 1
58. If a,b,c,d are odd natural numbers such that a + b + c + d = 20, C n 1 Cn 1
then the number of values of (a, b, c, d) is :
n 2  n 2
(a) 165 (b) 455 (c) 2 (d) None of these
C n 1
(c) 310 (d) 255
Ans. (c)
Ans. (a)
Sol. Given problem can be written as
Sol. Let a  2n1  1, b  2n2  1, c  2n3  1, d  2n4  1
x1  x 2  x 3 ..  x n  n 2 .
Now , given a  b  c  d  20
Such that x i  i, i  1 to n
 2  n1  n 2  n 7  n 4   4  20
Let x i  i=yi
 n1  n 2  n 3  n 4  8
since x i  i yi  0
Distribution of 8 identical objects in n1 , n 2 , n 3 , n 4 is
 equation (i) becomes :
8  4 1
C4 1 as n1 , n 2 , n 3 , n 4  0
 y1  1   y2  2    y3  3    y n  n   n 2 , yi  0
 C3  165
11

n(n  1)
59. Number of ways in which 25 identical balls can be distributed   y1  y2  y3   yn    n2
among Ram, shyam, Sunder and Ghanshyam such that at 2
least 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and
Ghanshyam respectively, is :  n2  n 
  y1  y2  y3   yn   n 2   
 2 
(a) 18 C4 (b) 28
C3
n2  n
(c) 24
C3 (d) 18
C3   y1  y2  y3   yn  
2
Ans. (d)
 Total no. of possibilities is given by
Sol. x1  x 2  x 3  x 4  25  (i)
n2  n n2  n  2 n  2 n2  n  2
 n 1
Let x1  1, x 2  2, x 3  3, x 4  4
2 Cn 1  2 Cn 1  2 Cn 1

Let x1  u1  1, x 2  u 2  2, x 3  u 3  3

x4  u4  4

from (i)