Maths-
If 
for 
, then 
Maths-General
- 0
- 1
- 2
- 3
Answer:The correct answer is: 1
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maths-
The largest value of r satisfying inequality 20Cr 
20Cr – 1 is -
The largest value of r satisfying inequality 20Cr 20Cr – 1 is -
maths-General
maths-
The maximum value of P such that 3P divides 99 × 97 × 95 × .... × 51 is-
99 × 97 × 95 ×....× 51
maximum power of 3 in 100 !
= 33 + 11 + 3 + 1 = 48.
maximum power of 3 in 50! =
= 16 + 5 + 1 = 22
maximum power of 3 in 25! =
= 8 + 2 = 10
exponent of 3 = 48 + 10 –
(22 × 2) = 14
maximum power of 3 in 100 !
= 33 + 11 + 3 + 1 = 48.
maximum power of 3 in 50! =
= 16 + 5 + 1 = 22
maximum power of 3 in 25! =
= 8 + 2 = 10
exponent of 3 = 48 + 10 –(22 × 2) = 14
The maximum value of P such that 3P divides 99 × 97 × 95 × .... × 51 is-
maths-General
99 × 97 × 95 ×....× 51
maximum power of 3 in 100 !
= 33 + 11 + 3 + 1 = 48.
maximum power of 3 in 50! =
= 16 + 5 + 1 = 22
maximum power of 3 in 25! =
= 8 + 2 = 10
exponent of 3 = 48 + 10 –
(22 × 2) = 14
maximum power of 3 in 100 !
= 33 + 11 + 3 + 1 = 48.
maximum power of 3 in 50! =
= 16 + 5 + 1 = 22
maximum power of 3 in 25! =
= 8 + 2 = 10
exponent of 3 = 48 + 10 –(22 × 2) = 14
maths-
A double decker bus can accommodate (u + 
) passengers, u in the upper deck and in the lower deck. The number of ways in which the (u + ) passengers can be distributed in the two decks, if r (
) particular passengers refuse to go in the upper deck and S(
u) refuse to sit in the lower deck, is -
A double decker bus can accommodate (u + ) passengers, u in the upper deck and in the lower deck. The number of ways in which the (u + ) passengers can be distributed in the two decks, if r () particular passengers refuse to go in the upper deck and S(u) refuse to sit in the lower deck, is -
maths-General
maths-
The number of five-digit telephone numbers having at least one of their digits repeated is
The number of five-digit telephone numbers which can be formed using the digits 0, 1, 2 ….9 is 105 . The number of five-digit telephone number which have none of their digits repeated is 10P5 = 30240. Thus, the required number of telephone number is 105 – 30240 = 69760.
The number of five-digit telephone numbers having at least one of their digits repeated is
maths-General
The number of five-digit telephone numbers which can be formed using the digits 0, 1, 2 ….9 is 105 . The number of five-digit telephone number which have none of their digits repeated is 10P5 = 30240. Thus, the required number of telephone number is 105 – 30240 = 69760.
maths-
One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways of doing this is -
Let the children be A,B.C
A B C
→19
Total number of ways
By termination
Permutation
A B C
→19Total number of ways
By termination
Permutation
One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways of doing this is -
maths-General
Let the children be A,B.C
A B C→19
Total number of ways
By termination
Permutation
A B C
→19Total number of ways
By termination
Permutation
maths-
Domain of f(x) = log10(log10(1 + x3)) is -
log10(1 + x3) > 0 Take antilog
1+ x3 > 1
x3 > 0 x > 0x
(0, 
) or x R+Domain of f(x) = log10(log10(1 + x3)) is -
maths-General
log10(1 + x3) > 0 Take antilog
1+ x3 > 1
x3 > 0 x > 0x
(0, ) or x R+maths-
Number of ways in which 25 identical balls can be distributed among Ram, Shyam, Sunder and Ghanshyam such that atleast 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and Ghanshyam respectively, is-
First of all select 1 + 2 + 3 + 4 = 10 balls out of 25 identical balls and distribute them as desired. It can happen only in one way. Now let x1, x2, x3 and x4 balls are given to them respectively.
(where x1, x2, x3, x4
0)
Its now one can get any number of balls
non negative integral solution of
x1 + x2 + x3 + x4 = 15
will be the number of ways so
15 + 4 –1C4–1 = 18C3
(where x1, x2, x3, x4
0)Its now one can get any number of balls
non negative integral solution ofx1 + x2 + x3 + x4 = 15
will be the number of ways so
15 + 4 –1C4–1 = 18C3
Number of ways in which 25 identical balls can be distributed among Ram, Shyam, Sunder and Ghanshyam such that atleast 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and Ghanshyam respectively, is-
maths-General
First of all select 1 + 2 + 3 + 4 = 10 balls out of 25 identical balls and distribute them as desired. It can happen only in one way. Now let x1, x2, x3 and x4 balls are given to them respectively.
(where x1, x2, x3, x4 0)
Its now one can get any number of balls
non negative integral solution of
x1 + x2 + x3 + x4 = 15
will be the number of ways so
15 + 4 –1C4–1 = 18C3
(where x1, x2, x3, x4
0)Its now one can get any number of balls
non negative integral solution ofx1 + x2 + x3 + x4 = 15
will be the number of ways so
15 + 4 –1C4–1 = 18C3
maths-
A seven digit number is in form of abcdefg (g, f, e, etc. are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are
Cases : i) If d = 6 then seven digit numbers possible are = 5C3. 3C3
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
ii) If d = 7 then numbers possible =,
iii) If d = 8 then numbers possible =
iv) If d = 9 then numbers possible =
Add all cases
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
ii) If d = 7 then numbers possible =,
iii) If d = 8 then numbers possible =
iv) If d = 9 then numbers possible =
Add all cases
A seven digit number is in form of abcdefg (g, f, e, etc. are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are
maths-General
Cases : i) If d = 6 then seven digit numbers possible are = 5C3. 3C3
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
ii) If d = 7 then numbers possible =,
iii) If d = 8 then numbers possible =
iv) If d = 9 then numbers possible =
Add all cases
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
ii) If d = 7 then numbers possible =,
iii) If d = 8 then numbers possible =
iv) If d = 9 then numbers possible =
Add all cases
maths-
Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy
5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.So this can be done is
= 10 × 3 × 2 × 4 = 240
Now one child can be rejected is
= 5 waysTotal ways = 5 × 240 = 1200Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy
maths-General
5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.So this can be done is
= 10 × 3 × 2 × 4 = 240
Now one child can be rejected is
= 5 waysTotal ways = 5 × 240 = 1200maths-
The total numbers of integral solutions for (x,y,z) such that xyz = 24 is
24 can be broken as (1, 1, 24),(1, 2, 12),(1, 3, 8),(1, 4, 6),(2, 3, 4) & (2, 2, 6)
All sets in which each no. is diff = 3C2. 3! + 3!= 24
in which two nos. are same =
total no. of possible sets= 24 × 4 + 12 × 2 = 120
All sets in which each no. is diff = 3C2. 3! + 3!= 24
in which two nos. are same =
total no. of possible sets= 24 × 4 + 12 × 2 = 120
The total numbers of integral solutions for (x,y,z) such that xyz = 24 is
maths-General
24 can be broken as (1, 1, 24),(1, 2, 12),(1, 3, 8),(1, 4, 6),(2, 3, 4) & (2, 2, 6)
All sets in which each no. is diff = 3C2. 3! + 3!= 24
in which two nos. are same =
total no. of possible sets= 24 × 4 + 12 × 2 = 120
All sets in which each no. is diff = 3C2. 3! + 3!= 24
in which two nos. are same =
total no. of possible sets= 24 × 4 + 12 × 2 = 120
maths-
A person writes letters to 6 friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least 4 of them are in wrong envelopes ?
no. of ways = 
= 135 + 264 + 265 = 664
= 135 + 264 + 265 = 664
A person writes letters to 6 friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least 4 of them are in wrong envelopes ?
maths-General
no. of ways =
= 135 + 264 + 265 = 664
= 135 + 264 + 265 = 664
maths-
Digit at unit place of sum (1!)2 +(2!)2 +(3!)2 ……… +(2008!)2 is
S = 1 + 4 + 36 + 576 + …….. +(2008!)2= 617 + all other terms will have 0 at unit placeDigit at unit place of sum (1!)2 +(2!)2 +(3!)2 ……… +(2008!)2 is
maths-General
S = 1 + 4 + 36 + 576 + …….. +(2008!)2= 617 + all other terms will have 0 at unit placemaths-
Number of rectangles in fig. shown which are not squares is
Total number of rectangles
No.of squares
No. of squares
Number of squares 
No.of squares No. of squares Number of squares Number of rectangles in fig. shown which are not squares is
maths-General
Total number of rectangles
No.of squares
No. of squares
Number of squares
No.of squares No. of squares Number of squares chemistry-
The correct orders about compounds I and II are:
i) 
ii) 
The correct orders about compounds I and II are:
i)
ii)
chemistry-General
chemistry-
Which of the following is correct method for separating a mixture of following compounds?
Which of the following is correct method for separating a mixture of following compounds?
chemistry-General