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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 14 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 14
What is the hundreds digits of ( 20 !-15!) ?
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0%
0
0%
1
0%
2
0%
4
0%
5
(
51
3
)
+
(
50
3
)
+
(
49
3
)
+
(
48
3
)
+
(
47
3
)
+
(
47
4
)
is equal to (where
(
n
r
)
denotes
n
C
r
Report Question
0%
(
52
1
)
0%
(
52
2
)
0%
(
52
3
)
0%
(
52
4
)
If
n
and
r
are positive integers such that
r
<
n
then
n
C
r
+
n
C
r
−
1
is equal to
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0%
2
n
C
2
r
−
1
0%
n
+
1
C
r
0%
n
C
r
+
1
0%
2
n
C
r
If
x
ϵ
N
a
n
d
C
x
−
1
4
−
C
X
−
1
3
−
5
4
P
x
−
2
2
<
0
then x is equal to
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0%
4,5,6,7,8,9,10
0%
3
0%
3
≤
x
≤
10
0%
none of these
If
m
and
n
are positive integers more than or equal to
2
,
m
>
n
,
then
(
m
n
)
!
is divisible by
Report Question
0%
(
m
!
)
n
0%
(
n
!
)
m
0%
(
m
+
n
)
!
0%
(
m
−
n
)
!
If
n
C
r
denotes the number of combinations of n things taken r at a time, then the expression +
n
C
r
−
1
+
2
×
n
C
r
equals
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0%
n
+
2
C
r
0%
n
+
2
C
r
+
1
0%
n
+
2
C
r
0%
n
+
2
C
r
+
1
If
189
C
35
+
189
C
x
=
190
C
x
, then
x
is equal to
Report Question
0%
34
0%
35
0%
36
0%
37
In a village, there are
87
families of which
52
families have at most
2
children. In a rural development programme,
20
families are to be helped chosen for assistance, of which at least
18
families must have at most
2
children. In how many ways can the choice be made?
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0%
52
C
18
×
35
C
2
+
52
C
19
×
35
C
1
+
52
C
20
×
35
C
0
0%
52
C
18
×
35
C
2
+
52
C
19
×
35
C
1
+
52
C
20
×
35
C
10
0%
52
C
18
×
35
C
2
+
52
C
19
×
35
C
1
+
52
C
2
×
5
C
0
0%
52
C
18
×
35
C
2
+
52
C
9
×
35
C
10
+
52
C
20
×
5
C
0
A sports team of
11
students is to be constituted, choosing at least
5
from class
X
I
and at least
5
from class
X
I
I
. If there are
20
students in each of these classes, in how many ways can the teams be constituted?
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0%
2
(
20
C
3
×
20
C
7
)
0%
2
(
20
C
5
×
20
C
6
)
0%
2
(
20
C
4
×
20
C
8
)
0%
None of these
Explanation
A team of
11
students can be constituted in the following two ways
i
)
5
student from class
X
I
and
6
from
X
I
I
i
i
)
6
student from class
X
I
and
5
from
X
I
I
∴
The required number of ways
⇒
20
C
5
×
20
C
6
+
20
C
6
×
20
C
5
⇒
2
(
20
C
5
×
20
C
6
)
The number of five digit numbers in which digits decrease from left to right, is
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0%
9
C
5
0%
10
C
5
0%
2
×
10
C
5
0%
10
P
5
If
n
C
r
+
4
n
C
r
+
1
+
6
n
C
r
+
2
+
4
n
C
r
+
3
+
n
C
r
+
4
n
C
r
+
3
n
C
r
+
1
+
3
n
C
r
+
2
+
n
C
r
+
3
=
n
+
k
r
+
k
then the value of
k
is
Report Question
0%
1
0%
2
0%
4
0%
5
The number of four -digit numbers strictly grater than
4321
that can be formed using the digits
0
,
1
,
2
,
3
,
4
,
5
(repetition of digits is allowed) is:
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0%
288
0%
306
0%
360
0%
310
Explanation
(1) The number of four-digit numbers Strating with 5 is equal to
6
3
=
216
(2) Starting with 44 and 55 is equal to
36
×
2
=
72
(3) Starting with 433, 434 and 435 is equal to
6
×
3
=
18
(4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
716
+
72
+
18
+
4
=
310
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