QBManager

Permutation and Combination

118981 The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is

1 271

2 272

3 300

4 None of these

Explanation:

C
Any number between 1 to 999 is of form the abc where, $0 \leq a, b, c \leq 9$ .
Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in $1 \times^{3} C_{1}$ $\times 9 \times 9 = 243$ ways, next 5 can occur in exactly two places in $^{3} C_{2} \times 9 = 27$
Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs
$= 1 \times 243 + 27 \times 2 + 1 \times 3$
$= 243 + 54 + 3$
$= 300$

BITSAT-2017

Permutation and Combination

118982 If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is

1 15

2 25

3 35

4 40

Explanation:

C Given that,
Number of identical pencil $= 16$
Number of children $= 4$
If each children gets at least 3 pencils then $3 \times 4 = 12$
pencils
Left pencil $= 16 - 12 = 4$
$n$ number of pencils can be distributed among $r$
persons $=^{n + r - 1} C_{r - 1}$
$=^{4 + 4 - 1} C_{4 - 1} =^{7} C_{3}$
$= \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5}{6} = 35$

BITSAT-2020

Permutation and Combination

118983 The number of possible outcomes in a throw of $n$ ordinary dice in which at least one of the dice shows an odd number is

1

6^{n} - 1

2

3^{n} - 1

3

6^{n} - 3^{n}

4 None of these

Explanation:

C We know that, there are total 6 face in a dice so the total number of ways $= 6 \times 6 \times \dots . n = 6^{n}$
Showing face in every time $= 3$
So, the total number of ways to show only even number $= 3 \times 3 \times \dots \dots n = 3^{n}$
$∴$ Required number of ways $=$ Total face-showing face
$= 6^{n} - 3^{n}$

BITSAT - 2020

Permutation and Combination

118984 A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

1

\frac{(10)!}{6!}

2

\frac{(10)!}{24}

3

\frac{(9)!}{24}

4 None of these

Explanation:

B Ways to select 6 people out of 10 people\)
$^{10} C_{6}$
Permutation of 6 on round table $= 5!$
Ways to arrange four people at the first $= (4 - 1)!$
| Permutation of 4 on round table $= 3$ !
Then, total number of arrangements $=^{10} C_{6} .5! .3$ !
$= \frac{(10)!}{6! .4!} \cdot 5! 3! = \frac{(10)!}{24} .$

BITSAT-2020

Permutation and Combination

118981 The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is

1 271

2 272

3 300

4 None of these

Explanation:

C
Any number between 1 to 999 is of form the abc where, $0 \leq a, b, c \leq 9$ .
Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in $1 \times^{3} C_{1}$ $\times 9 \times 9 = 243$ ways, next 5 can occur in exactly two places in $^{3} C_{2} \times 9 = 27$
Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs
$= 1 \times 243 + 27 \times 2 + 1 \times 3$
$= 243 + 54 + 3$
$= 300$

BITSAT-2017

Permutation and Combination

118982 If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is

1 15

2 25

3 35

4 40

Explanation:

C Given that,
Number of identical pencil $= 16$
Number of children $= 4$
If each children gets at least 3 pencils then $3 \times 4 = 12$
pencils
Left pencil $= 16 - 12 = 4$
$n$ number of pencils can be distributed among $r$
persons $=^{n + r - 1} C_{r - 1}$
$=^{4 + 4 - 1} C_{4 - 1} =^{7} C_{3}$
$= \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5}{6} = 35$

BITSAT-2020

Permutation and Combination

118983 The number of possible outcomes in a throw of $n$ ordinary dice in which at least one of the dice shows an odd number is

1

6^{n} - 1

2

3^{n} - 1

3

6^{n} - 3^{n}

4 None of these

Explanation:

C We know that, there are total 6 face in a dice so the total number of ways $= 6 \times 6 \times \dots . n = 6^{n}$
Showing face in every time $= 3$
So, the total number of ways to show only even number $= 3 \times 3 \times \dots \dots n = 3^{n}$
$∴$ Required number of ways $=$ Total face-showing face
$= 6^{n} - 3^{n}$

BITSAT - 2020

Permutation and Combination

118984 A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

1

\frac{(10)!}{6!}

2

\frac{(10)!}{24}

3

\frac{(9)!}{24}

4 None of these

Explanation:

B Ways to select 6 people out of 10 people\)
$^{10} C_{6}$
Permutation of 6 on round table $= 5!$
Ways to arrange four people at the first $= (4 - 1)!$
| Permutation of 4 on round table $= 3$ !
Then, total number of arrangements $=^{10} C_{6} .5! .3$ !
$= \frac{(10)!}{6! .4!} \cdot 5! 3! = \frac{(10)!}{24} .$

BITSAT-2020

Permutation and Combination

118981 The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is

1 271

2 272

3 300

4 None of these

Explanation:

C
Any number between 1 to 999 is of form the abc where, $0 \leq a, b, c \leq 9$ .
Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in $1 \times^{3} C_{1}$ $\times 9 \times 9 = 243$ ways, next 5 can occur in exactly two places in $^{3} C_{2} \times 9 = 27$
Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs
$= 1 \times 243 + 27 \times 2 + 1 \times 3$
$= 243 + 54 + 3$
$= 300$

BITSAT-2017

Permutation and Combination

118982 If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is

1 15

2 25

3 35

4 40

Explanation:

C Given that,
Number of identical pencil $= 16$
Number of children $= 4$
If each children gets at least 3 pencils then $3 \times 4 = 12$
pencils
Left pencil $= 16 - 12 = 4$
$n$ number of pencils can be distributed among $r$
persons $=^{n + r - 1} C_{r - 1}$
$=^{4 + 4 - 1} C_{4 - 1} =^{7} C_{3}$
$= \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5}{6} = 35$

BITSAT-2020

Permutation and Combination

118983 The number of possible outcomes in a throw of $n$ ordinary dice in which at least one of the dice shows an odd number is

1

6^{n} - 1

2

3^{n} - 1

3

6^{n} - 3^{n}

4 None of these

Explanation:

C We know that, there are total 6 face in a dice so the total number of ways $= 6 \times 6 \times \dots . n = 6^{n}$
Showing face in every time $= 3$
So, the total number of ways to show only even number $= 3 \times 3 \times \dots \dots n = 3^{n}$
$∴$ Required number of ways $=$ Total face-showing face
$= 6^{n} - 3^{n}$

BITSAT - 2020

Permutation and Combination

118984 A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

1

\frac{(10)!}{6!}

2

\frac{(10)!}{24}

3

\frac{(9)!}{24}

4 None of these

Explanation:

B Ways to select 6 people out of 10 people\)
$^{10} C_{6}$
Permutation of 6 on round table $= 5!$
Ways to arrange four people at the first $= (4 - 1)!$
| Permutation of 4 on round table $= 3$ !
Then, total number of arrangements $=^{10} C_{6} .5! .3$ !
$= \frac{(10)!}{6! .4!} \cdot 5! 3! = \frac{(10)!}{24} .$

BITSAT-2020

Permutation and Combination

118981 The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is

1 271

2 272

3 300

4 None of these

Explanation:

C
Any number between 1 to 999 is of form the abc where, $0 \leq a, b, c \leq 9$ .
Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in $1 \times^{3} C_{1}$ $\times 9 \times 9 = 243$ ways, next 5 can occur in exactly two places in $^{3} C_{2} \times 9 = 27$
Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs
$= 1 \times 243 + 27 \times 2 + 1 \times 3$
$= 243 + 54 + 3$
$= 300$

BITSAT-2017

Permutation and Combination

118982 If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is

1 15

2 25

3 35

4 40

Explanation:

C Given that,
Number of identical pencil $= 16$
Number of children $= 4$
If each children gets at least 3 pencils then $3 \times 4 = 12$
pencils
Left pencil $= 16 - 12 = 4$
$n$ number of pencils can be distributed among $r$
persons $=^{n + r - 1} C_{r - 1}$
$=^{4 + 4 - 1} C_{4 - 1} =^{7} C_{3}$
$= \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5}{6} = 35$

BITSAT-2020

Permutation and Combination

118983 The number of possible outcomes in a throw of $n$ ordinary dice in which at least one of the dice shows an odd number is

1

6^{n} - 1

2

3^{n} - 1

3

6^{n} - 3^{n}

4 None of these

Explanation:

C We know that, there are total 6 face in a dice so the total number of ways $= 6 \times 6 \times \dots . n = 6^{n}$
Showing face in every time $= 3$
So, the total number of ways to show only even number $= 3 \times 3 \times \dots \dots n = 3^{n}$
$∴$ Required number of ways $=$ Total face-showing face
$= 6^{n} - 3^{n}$

BITSAT - 2020

Permutation and Combination

118984 A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

1

\frac{(10)!}{6!}

2

\frac{(10)!}{24}

3

\frac{(9)!}{24}

4 None of these

Explanation:

B Ways to select 6 people out of 10 people\)
$^{10} C_{6}$
Permutation of 6 on round table $= 5!$
Ways to arrange four people at the first $= (4 - 1)!$
| Permutation of 4 on round table $= 3$ !
Then, total number of arrangements $=^{10} C_{6} .5! .3$ !
$= \frac{(10)!}{6! .4!} \cdot 5! 3! = \frac{(10)!}{24} .$

BITSAT-2020

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