Maths-

General

Easy

### Question

#### The total numbers of integral solutions for (x,y,z) such that xyz = 24 is

- 36
- 90
- 96
- 120

#### The correct answer is: 120

#### 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 = ^{3}C_{2}. 3! + 3!= 24

in which two nos. are same =

total no. of possible sets= 24 × 4 + 12 × 2 = 120

## Book A Free Demo

+91

Grade*

Select Grade

### Related Questions to study

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 ways

Total ways = 5 × 240 = 1200

So this can be done is

= 10 × 3 × 2 × 4 = 240

Now one child can be rejected is = 5 ways

Total ways = 5 × 240 = 1200

#### Number 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 ways

Total ways = 5 × 240 = 1200

So this can be done is

= 10 × 3 × 2 × 4 = 240

Now one child can be rejected is = 5 ways

Total ways = 5 × 240 = 1200

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 =

[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

^{5}C_{3}.^{3}C_{3}[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 =

[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

^{5}C_{3}.^{3}C_{3}[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 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 x

(where x

Its now one can get any number of balls

non negative integral solution of

x

will be the number of ways so

_{1}, x_{2}, x_{3}and x_{4}balls are given to them respectively.(where x

_{1}, x_{2}, x_{3}, x_{4}0)Its now one can get any number of balls

non negative integral solution of

x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15will be the number of ways so

^{15 + 4 –1}C_{4–1}=^{18}C_{3}#### 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 x

(where x

Its now one can get any number of balls

non negative integral solution of

x

will be the number of ways so

_{1}, x_{2}, x_{3}and x_{4}balls are given to them respectively.(where x

_{1}, x_{2}, x_{3}, x_{4}0)Its now one can get any number of balls

non negative integral solution of

x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15will be the number of ways so

^{15 + 4 –1}C_{4–1}=^{18}C_{3}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 →19

Total 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 →19

Total number of ways

By termination

Permutation

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 10

^{5}. The number of five-digit telephone number which have none of their digits repeated is^{10}P_{5}= 30240. Thus, the required number of telephone number is 10^{5}– 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 10

^{5}. The number of five-digit telephone number which have none of their digits repeated is^{10}P_{5}= 30240. Thus, the required number of telephone number is 10^{5}– 30240 = 69760.maths-

#### The maximum value of P such that 3^{P} 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 3^{P} 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-

#### If for , then

#### If for , then

maths-General

maths-

#### The largest value of r satisfying inequality ^{20}C_{r }^{20}C_{r}_{ – 1 }is -

#### The largest value of r satisfying inequality ^{20}C_{r }^{20}C_{r}_{ – 1 }is -

maths-General

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-

#### Domain of f(x) = log_{10}(log_{10}(1 + x^{3})) is -

log

(1 + x

1+ x

x (0, ) or x R

_{10}(1 + x

^{3}) > 0 Take antilog1+ x

^{3}> 1 x^{3}> 0 x > 0x (0, ) or x R

^{+}#### Domain of f(x) = log_{10}(log_{10}(1 + x^{3})) is -

maths-General

log

(1 + x

1+ x

x (0, ) or x R

_{10}(1 + x

^{3}) > 0 Take antilog1+ x

^{3}> 1 x^{3}> 0 x > 0x (0, ) or x R

^{+}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

chemistry-

#### The order in which the reagent must be used to separate the compound I - IV is:

#### The order in which the reagent must be used to separate the compound I - IV is:

chemistry-General

chemistry-

#### Decreasing order of solubility of following compounds is:

i)

ii)

iii)

iv)

#### Decreasing order of solubility of following compounds is:

i)

ii)

iii)

iv)

chemistry-General

chemistry-

#### Which of the following statement is correct about tropolone?

#### Which of the following statement is correct about tropolone?

chemistry-General