Permutations & Combinations Practice Questions - DPP for JEE

 Page 2


the vertices of a regular polygon of n sides.
? T
n
 = 
n
C
3
 ? T
n+1
 = 
n+1
C
3
? T
n+1
 – T
n
 = 
n+1
C
3
 – 
n
C
3
 = 28 (Given)
? 
?  ? n(n – 1) = 28 × 2
? (n – 8) (n + 7) = 0 ? n = 8, – 7
n can never be less than zero
? n = 8
5. (a)
x+2
P
x+2
 = a ? a = ( x +2 )!
x
P
11
 = b ? b =  and   
x–11
P
x–11
 = c
? c = (x – 11)!
a = 182 bc
? (x + 2)!  = 182  (x– 11)!
? (x + 2) (x + 1) = 182 = 14 × 13
? x + 1 = 13 ? x = 12
6. (a) We know that in any triangle the sum of two sides is always
greater than the third side.
? The triangle will not be formed if we select segments of length (2,
3, 5), (2, 3, 6) or (2, 4, 6).
Hence no. of triangles formed = 
5
C
3
 – 3.
7. (c) There are 3p points, so possible no. of  triangles = 
3p
C
3
But the points on he same line do not form a triangle, such selections are
= 3 . 
p
C
3
?  Desired number = 
3p
C
3
 – 3 . 
p
C
3
 = p
2
 (4p – 3)
8. (c) Leaving the ground floor and second floor, their are 10 floors in
which three groups of people can left the lift cabin in 
10
P
3
 ways, i.e.
720 ways.
Page 3


the vertices of a regular polygon of n sides.
? T
n
 = 
n
C
3
 ? T
n+1
 = 
n+1
C
3
? T
n+1
 – T
n
 = 
n+1
C
3
 – 
n
C
3
 = 28 (Given)
? 
?  ? n(n – 1) = 28 × 2
? (n – 8) (n + 7) = 0 ? n = 8, – 7
n can never be less than zero
? n = 8
5. (a)
x+2
P
x+2
 = a ? a = ( x +2 )!
x
P
11
 = b ? b =  and   
x–11
P
x–11
 = c
? c = (x – 11)!
a = 182 bc
? (x + 2)!  = 182  (x– 11)!
? (x + 2) (x + 1) = 182 = 14 × 13
? x + 1 = 13 ? x = 12
6. (a) We know that in any triangle the sum of two sides is always
greater than the third side.
? The triangle will not be formed if we select segments of length (2,
3, 5), (2, 3, 6) or (2, 4, 6).
Hence no. of triangles formed = 
5
C
3
 – 3.
7. (c) There are 3p points, so possible no. of  triangles = 
3p
C
3
But the points on he same line do not form a triangle, such selections are
= 3 . 
p
C
3
?  Desired number = 
3p
C
3
 – 3 . 
p
C
3
 = p
2
 (4p – 3)
8. (c) Leaving the ground floor and second floor, their are 10 floors in
which three groups of people can left the lift cabin in 
10
P
3
 ways, i.e.
720 ways.
9. (d) There are in the set (1, 2, 3, ..... n) (n being odd),  even
numbers,  odd numbers and for an A.P., the sum of the
extremes is always even and hence the choice is either both even
or both odd and this may be done in
 ways
Note that, if a, b, c are in A.P. a + c = 2b. Hence, if a, b, c are integer the
sum of extreme digits (a and c) is even.
10. (b) Total number of possible predictions = 3
10
No. of predictions which have r wrong and 10 – r correct entries = 
10
C
r
2
10 – r
? Desired no. of ways = 3
10
 – 
[Note that above no. of ways is also equal to 
11. (d) A chessboard is made up of 9 equispaced horizontal and vertical
line. To make a 1 × 1 square, we must choose two consecutive
horizontal and vertical lines from among these. This can be done in
8 × 8 = 8
2
 ways. A 2 × 2 square needs three consecutive horizontal
and vertical lines, and we can do this in 7 × 7 = 7
2
 ways.
Continuing in this manner, the total number of square is
= 204.
12. (d) Number of all possible triangles
= Number of selections of 3 points from 8 vertices = 
8
C
3
 = 56
Page 4


the vertices of a regular polygon of n sides.
? T
n
 = 
n
C
3
 ? T
n+1
 = 
n+1
C
3
? T
n+1
 – T
n
 = 
n+1
C
3
 – 
n
C
3
 = 28 (Given)
? 
?  ? n(n – 1) = 28 × 2
? (n – 8) (n + 7) = 0 ? n = 8, – 7
n can never be less than zero
? n = 8
5. (a)
x+2
P
x+2
 = a ? a = ( x +2 )!
x
P
11
 = b ? b =  and   
x–11
P
x–11
 = c
? c = (x – 11)!
a = 182 bc
? (x + 2)!  = 182  (x– 11)!
? (x + 2) (x + 1) = 182 = 14 × 13
? x + 1 = 13 ? x = 12
6. (a) We know that in any triangle the sum of two sides is always
greater than the third side.
? The triangle will not be formed if we select segments of length (2,
3, 5), (2, 3, 6) or (2, 4, 6).
Hence no. of triangles formed = 
5
C
3
 – 3.
7. (c) There are 3p points, so possible no. of  triangles = 
3p
C
3
But the points on he same line do not form a triangle, such selections are
= 3 . 
p
C
3
?  Desired number = 
3p
C
3
 – 3 . 
p
C
3
 = p
2
 (4p – 3)
8. (c) Leaving the ground floor and second floor, their are 10 floors in
which three groups of people can left the lift cabin in 
10
P
3
 ways, i.e.
720 ways.
9. (d) There are in the set (1, 2, 3, ..... n) (n being odd),  even
numbers,  odd numbers and for an A.P., the sum of the
extremes is always even and hence the choice is either both even
or both odd and this may be done in
 ways
Note that, if a, b, c are in A.P. a + c = 2b. Hence, if a, b, c are integer the
sum of extreme digits (a and c) is even.
10. (b) Total number of possible predictions = 3
10
No. of predictions which have r wrong and 10 – r correct entries = 
10
C
r
2
10 – r
? Desired no. of ways = 3
10
 – 
[Note that above no. of ways is also equal to 
11. (d) A chessboard is made up of 9 equispaced horizontal and vertical
line. To make a 1 × 1 square, we must choose two consecutive
horizontal and vertical lines from among these. This can be done in
8 × 8 = 8
2
 ways. A 2 × 2 square needs three consecutive horizontal
and vertical lines, and we can do this in 7 × 7 = 7
2
 ways.
Continuing in this manner, the total number of square is
= 204.
12. (d) Number of all possible triangles
= Number of selections of 3 points from 8 vertices = 
8
C
3
 = 56
Number of triangle with one side common with octagon
=  8 × 4 = 32
(Consider side A
1
A
2
. Since two points A
3
, A
8
 are adjacent, 3rd point
should be chosen from remaining 4 points.)
Number of triangles having two sides common with octagon : All such
triangles have three consecutive vertices, viz., A
1
A
2
A
3
, A
2
A
3
A
4
,
..... A
8
A
1
A
2
.
Number of such triangles = 8
? Number of triangles with no side common
= 56 – 32 – 8 = 16.
13. (d) When A has B or C to his right we have the order :
AB or AC ...(1)
When B has C or D to his right, we have the order :
BC or BD ...(2)
Taking these two possibilities together, we must have ABC or ABD or
AC and BD.
For ABC, D, E, F to arrange along a circle, number of way = 3 ! = 6,
where three persons A,B,C together are treated as single.
For ABD, C, E, F, the number of ways = 6. For AC, BD, E,F the number
of ways = 6.
Hence, total number of ways = 18.
14. (c) Let m + 5 
Pm + 1
 = 
? = 
?  = 
(m + 4) (m + 5) = 22(m – 1)
Page 5


the vertices of a regular polygon of n sides.
? T
n
 = 
n
C
3
 ? T
n+1
 = 
n+1
C
3
? T
n+1
 – T
n
 = 
n+1
C
3
 – 
n
C
3
 = 28 (Given)
? 
?  ? n(n – 1) = 28 × 2
? (n – 8) (n + 7) = 0 ? n = 8, – 7
n can never be less than zero
? n = 8
5. (a)
x+2
P
x+2
 = a ? a = ( x +2 )!
x
P
11
 = b ? b =  and   
x–11
P
x–11
 = c
? c = (x – 11)!
a = 182 bc
? (x + 2)!  = 182  (x– 11)!
? (x + 2) (x + 1) = 182 = 14 × 13
? x + 1 = 13 ? x = 12
6. (a) We know that in any triangle the sum of two sides is always
greater than the third side.
? The triangle will not be formed if we select segments of length (2,
3, 5), (2, 3, 6) or (2, 4, 6).
Hence no. of triangles formed = 
5
C
3
 – 3.
7. (c) There are 3p points, so possible no. of  triangles = 
3p
C
3
But the points on he same line do not form a triangle, such selections are
= 3 . 
p
C
3
?  Desired number = 
3p
C
3
 – 3 . 
p
C
3
 = p
2
 (4p – 3)
8. (c) Leaving the ground floor and second floor, their are 10 floors in
which three groups of people can left the lift cabin in 
10
P
3
 ways, i.e.
720 ways.
9. (d) There are in the set (1, 2, 3, ..... n) (n being odd),  even
numbers,  odd numbers and for an A.P., the sum of the
extremes is always even and hence the choice is either both even
or both odd and this may be done in
 ways
Note that, if a, b, c are in A.P. a + c = 2b. Hence, if a, b, c are integer the
sum of extreme digits (a and c) is even.
10. (b) Total number of possible predictions = 3
10
No. of predictions which have r wrong and 10 – r correct entries = 
10
C
r
2
10 – r
? Desired no. of ways = 3
10
 – 
[Note that above no. of ways is also equal to 
11. (d) A chessboard is made up of 9 equispaced horizontal and vertical
line. To make a 1 × 1 square, we must choose two consecutive
horizontal and vertical lines from among these. This can be done in
8 × 8 = 8
2
 ways. A 2 × 2 square needs three consecutive horizontal
and vertical lines, and we can do this in 7 × 7 = 7
2
 ways.
Continuing in this manner, the total number of square is
= 204.
12. (d) Number of all possible triangles
= Number of selections of 3 points from 8 vertices = 
8
C
3
 = 56
Number of triangle with one side common with octagon
=  8 × 4 = 32
(Consider side A
1
A
2
. Since two points A
3
, A
8
 are adjacent, 3rd point
should be chosen from remaining 4 points.)
Number of triangles having two sides common with octagon : All such
triangles have three consecutive vertices, viz., A
1
A
2
A
3
, A
2
A
3
A
4
,
..... A
8
A
1
A
2
.
Number of such triangles = 8
? Number of triangles with no side common
= 56 – 32 – 8 = 16.
13. (d) When A has B or C to his right we have the order :
AB or AC ...(1)
When B has C or D to his right, we have the order :
BC or BD ...(2)
Taking these two possibilities together, we must have ABC or ABD or
AC and BD.
For ABC, D, E, F to arrange along a circle, number of way = 3 ! = 6,
where three persons A,B,C together are treated as single.
For ABD, C, E, F, the number of ways = 6. For AC, BD, E,F the number
of ways = 6.
Hence, total number of ways = 18.
14. (c) Let m + 5 
Pm + 1
 = 
? = 
?  = 
(m + 4) (m + 5) = 22(m – 1)
? m
2
 – 13m + 42 = 0
? m
2
 – 7m – 6m + 42 = 0
? m(m – 7) – 6 (m – 7) = 0
? m = 6, 7
Hence  m
1
 + m
2
 = 13
15. (b) There are 8 chairs on each side of the table. Let the sides be
represented by A and B. Let four persons sit on side A, then
number of ways of arranging 4 persons on 8 chairs on side A = 
8
P
4
and then two persons sit on side B. The number of ways of
arranging 2 persons on 8 chairs on side B = 
8
P
2
 and the remaining
10 persons can be arranged in remaining 10 chairs in 10! ways.
Hence the total number of ways in which the persons can be arranged
= 
8
P
4
 × 
8
P
2
? 
× 10! = 
16. (d) Number of white balls = 10
Number of green balls = 9
and number of black balls = 7
? Required probability = (10 + 1) (9 + 1) (7 + 1) – 1
= 11.10.8 –1 = 879
17. (a) Any number greater than a million must be of 7 or more than 7
digits. Here number of given digits is seven, therefoer we have to
form numbers of seven digits only.
Now there are seven digits of which 3 occurs thrice and two occurs
twice.
? number of numbers formed =  = 420
But this also includes those numbers of seven digits whose first digit is
zero and so in fact they are only six digit numbers.
Number of numbers of seven digits having zero in the first place = 1 × 
 = 60.
Hence required number = 420 – 60 = 360
18. (a) Let us first consider 2 letters and 2 envelopes, then there is only
one way to place both the letters in wrong envelope.