QBManager

Permutation and Combination

119216 Six persons A, B, C, D, E and $F$ are to be seated at a circular table facing towards the centre. Then the number of ways that can be done if $A$ must have either $E$ or $F$ on his immediate right and $E$ must have either $F$ or $D$ on his immediate right, is

1 18

2 30

3 12

4 24

Explanation:

A Given that,
Number of Person $= 6$
They are sitting around a circular table facing toward
the centre.
Here $B$ sits on right side of $A$ ,
$D$ or $C$ will be right of $B$ .
This can be done in 2 ! ways.
Remaining three place can be filled in 3 ! ways
$= 2$ ! $\times 3$ ! $= 2 \times 1 \times 3 \times 2 \times 1 = 12$ ways
Say $C$ on right of 3 places.
B immediately next is $D$
A, B, C, D are filled in 3 ways
Remaining two places in 2 ways
Total $= 6$ ways
Total number $= 12 + 6 = 18$

AP EAMCET-21.04.2019

Permutation and Combination

119217 How many four-digit numbers abcd exist such that $a$ is odd, $b$ is divisible by $3, c$ is even and $d$ is prime?

1 380

2 360

3 400

4 480

Explanation:

C As we know that, the odd number be $(1, 3, 5$ , 7,9 )
Therefore $n (a) = 5$
Now, divisible by 3 are ${0, 3, 6, 9}, n (b) = 4$
For even number are ${0, 2, 4, 6, 8}, n (c) = 5$
And also the prime number are ${2, 3, 5, 7}, n (d) = 4$
Hence, the four digit number a $b d$ exist
$= 5 \times 4 \times 5 \times 4 = 25 \times 16 = 400$

AP EAMCET-05.10.2021

Permutation and Combination

119218 The remainder obtained when $(1!)^{2} + (2!)^{2} +$ $(3!)^{2} + \dots + (100!)^{2}$ is divided by $10^{2}$ is

1 27

2 28

3 17

4 14

Explanation:

C According to question,
$\frac{(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2} + (5!)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
$= \frac{1 + (2)^{2} + (6)^{2} + (24)^{2} + (120)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
Since, $(5!)^{2}, (6!)^{2}, (7!)^{2} \dots \dots (100!)^{2}$ each multiple of $(10)^{2}$
And for $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2}$
$∴ (1)^{2} + (2)^{2} + (6)^{2} + (24)^{2}$
$= 1 + 4 + 36 + 576$
$= 617$ When, 617 is divided by 100 its remainder is 17 .

COMEDK-2016

Permutation and Combination

119228 The English alphabets have 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed from the alphabet?

1

2100 \times 2

!

2

210 \times 2

!

3

210 \times 4

!

4

2100 \times 4

!

Explanation:

D 2 Vowels can be chosen in $^{5} C_{2}$ ways
And, 2 consonant can be chosen in $^{21} C_{2}$ ways.
4 letter can be arranged in 4 ways
$=^{5} C_{2} \times^{21} C_{2} \times 4$ !
$= \frac{5!}{2! (5 - 2)!} \times \frac{21!}{2! (21 - 2)!} \times 4$ !
$= \frac{5!}{2! 3!} \times \frac{21!}{2! 19!} \times 4$ !
$= \frac{5 \times 4 \times 3!}{2! 3!} \times \frac{21 \times 20 \times 19!}{2! \times 19!} \times 4$ !
$= 10 \times 210 \times 4! = 2100 \times 4$ !

AP EAMCET-07.07.2022

Permutation and Combination

119216 Six persons A, B, C, D, E and $F$ are to be seated at a circular table facing towards the centre. Then the number of ways that can be done if $A$ must have either $E$ or $F$ on his immediate right and $E$ must have either $F$ or $D$ on his immediate right, is

1 18

2 30

3 12

4 24

Explanation:

A Given that,
Number of Person $= 6$
They are sitting around a circular table facing toward
the centre.
Here $B$ sits on right side of $A$ ,
$D$ or $C$ will be right of $B$ .
This can be done in 2 ! ways.
Remaining three place can be filled in 3 ! ways
$= 2$ ! $\times 3$ ! $= 2 \times 1 \times 3 \times 2 \times 1 = 12$ ways
Say $C$ on right of 3 places.
B immediately next is $D$
A, B, C, D are filled in 3 ways
Remaining two places in 2 ways
Total $= 6$ ways
Total number $= 12 + 6 = 18$

AP EAMCET-21.04.2019

Permutation and Combination

119217 How many four-digit numbers abcd exist such that $a$ is odd, $b$ is divisible by $3, c$ is even and $d$ is prime?

1 380

2 360

3 400

4 480

Explanation:

C As we know that, the odd number be $(1, 3, 5$ , 7,9 )
Therefore $n (a) = 5$
Now, divisible by 3 are ${0, 3, 6, 9}, n (b) = 4$
For even number are ${0, 2, 4, 6, 8}, n (c) = 5$
And also the prime number are ${2, 3, 5, 7}, n (d) = 4$
Hence, the four digit number a $b d$ exist
$= 5 \times 4 \times 5 \times 4 = 25 \times 16 = 400$

AP EAMCET-05.10.2021

Permutation and Combination

119218 The remainder obtained when $(1!)^{2} + (2!)^{2} +$ $(3!)^{2} + \dots + (100!)^{2}$ is divided by $10^{2}$ is

1 27

2 28

3 17

4 14

Explanation:

C According to question,
$\frac{(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2} + (5!)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
$= \frac{1 + (2)^{2} + (6)^{2} + (24)^{2} + (120)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
Since, $(5!)^{2}, (6!)^{2}, (7!)^{2} \dots \dots (100!)^{2}$ each multiple of $(10)^{2}$
And for $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2}$
$∴ (1)^{2} + (2)^{2} + (6)^{2} + (24)^{2}$
$= 1 + 4 + 36 + 576$
$= 617$ When, 617 is divided by 100 its remainder is 17 .

COMEDK-2016

Permutation and Combination

119228 The English alphabets have 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed from the alphabet?

1

2100 \times 2

!

2

210 \times 2

!

3

210 \times 4

!

4

2100 \times 4

!

Explanation:

D 2 Vowels can be chosen in $^{5} C_{2}$ ways
And, 2 consonant can be chosen in $^{21} C_{2}$ ways.
4 letter can be arranged in 4 ways
$=^{5} C_{2} \times^{21} C_{2} \times 4$ !
$= \frac{5!}{2! (5 - 2)!} \times \frac{21!}{2! (21 - 2)!} \times 4$ !
$= \frac{5!}{2! 3!} \times \frac{21!}{2! 19!} \times 4$ !
$= \frac{5 \times 4 \times 3!}{2! 3!} \times \frac{21 \times 20 \times 19!}{2! \times 19!} \times 4$ !
$= 10 \times 210 \times 4! = 2100 \times 4$ !

AP EAMCET-07.07.2022

Permutation and Combination

119216 Six persons A, B, C, D, E and $F$ are to be seated at a circular table facing towards the centre. Then the number of ways that can be done if $A$ must have either $E$ or $F$ on his immediate right and $E$ must have either $F$ or $D$ on his immediate right, is

1 18

2 30

3 12

4 24

Explanation:

A Given that,
Number of Person $= 6$
They are sitting around a circular table facing toward
the centre.
Here $B$ sits on right side of $A$ ,
$D$ or $C$ will be right of $B$ .
This can be done in 2 ! ways.
Remaining three place can be filled in 3 ! ways
$= 2$ ! $\times 3$ ! $= 2 \times 1 \times 3 \times 2 \times 1 = 12$ ways
Say $C$ on right of 3 places.
B immediately next is $D$
A, B, C, D are filled in 3 ways
Remaining two places in 2 ways
Total $= 6$ ways
Total number $= 12 + 6 = 18$

AP EAMCET-21.04.2019

Permutation and Combination

119217 How many four-digit numbers abcd exist such that $a$ is odd, $b$ is divisible by $3, c$ is even and $d$ is prime?

1 380

2 360

3 400

4 480

Explanation:

C As we know that, the odd number be $(1, 3, 5$ , 7,9 )
Therefore $n (a) = 5$
Now, divisible by 3 are ${0, 3, 6, 9}, n (b) = 4$
For even number are ${0, 2, 4, 6, 8}, n (c) = 5$
And also the prime number are ${2, 3, 5, 7}, n (d) = 4$
Hence, the four digit number a $b d$ exist
$= 5 \times 4 \times 5 \times 4 = 25 \times 16 = 400$

AP EAMCET-05.10.2021

Permutation and Combination

119218 The remainder obtained when $(1!)^{2} + (2!)^{2} +$ $(3!)^{2} + \dots + (100!)^{2}$ is divided by $10^{2}$ is

1 27

2 28

3 17

4 14

Explanation:

C According to question,
$\frac{(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2} + (5!)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
$= \frac{1 + (2)^{2} + (6)^{2} + (24)^{2} + (120)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
Since, $(5!)^{2}, (6!)^{2}, (7!)^{2} \dots \dots (100!)^{2}$ each multiple of $(10)^{2}$
And for $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2}$
$∴ (1)^{2} + (2)^{2} + (6)^{2} + (24)^{2}$
$= 1 + 4 + 36 + 576$
$= 617$ When, 617 is divided by 100 its remainder is 17 .

COMEDK-2016

Permutation and Combination

119228 The English alphabets have 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed from the alphabet?

1

2100 \times 2

!

2

210 \times 2

!

3

210 \times 4

!

4

2100 \times 4

!

Explanation:

D 2 Vowels can be chosen in $^{5} C_{2}$ ways
And, 2 consonant can be chosen in $^{21} C_{2}$ ways.
4 letter can be arranged in 4 ways
$=^{5} C_{2} \times^{21} C_{2} \times 4$ !
$= \frac{5!}{2! (5 - 2)!} \times \frac{21!}{2! (21 - 2)!} \times 4$ !
$= \frac{5!}{2! 3!} \times \frac{21!}{2! 19!} \times 4$ !
$= \frac{5 \times 4 \times 3!}{2! 3!} \times \frac{21 \times 20 \times 19!}{2! \times 19!} \times 4$ !
$= 10 \times 210 \times 4! = 2100 \times 4$ !

AP EAMCET-07.07.2022

Permutation and Combination

119216 Six persons A, B, C, D, E and $F$ are to be seated at a circular table facing towards the centre. Then the number of ways that can be done if $A$ must have either $E$ or $F$ on his immediate right and $E$ must have either $F$ or $D$ on his immediate right, is

1 18

2 30

3 12

4 24

Explanation:

A Given that,
Number of Person $= 6$
They are sitting around a circular table facing toward
the centre.
Here $B$ sits on right side of $A$ ,
$D$ or $C$ will be right of $B$ .
This can be done in 2 ! ways.
Remaining three place can be filled in 3 ! ways
$= 2$ ! $\times 3$ ! $= 2 \times 1 \times 3 \times 2 \times 1 = 12$ ways
Say $C$ on right of 3 places.
B immediately next is $D$
A, B, C, D are filled in 3 ways
Remaining two places in 2 ways
Total $= 6$ ways
Total number $= 12 + 6 = 18$

AP EAMCET-21.04.2019

Permutation and Combination

119217 How many four-digit numbers abcd exist such that $a$ is odd, $b$ is divisible by $3, c$ is even and $d$ is prime?

1 380

2 360

3 400

4 480

Explanation:

C As we know that, the odd number be $(1, 3, 5$ , 7,9 )
Therefore $n (a) = 5$
Now, divisible by 3 are ${0, 3, 6, 9}, n (b) = 4$
For even number are ${0, 2, 4, 6, 8}, n (c) = 5$
And also the prime number are ${2, 3, 5, 7}, n (d) = 4$
Hence, the four digit number a $b d$ exist
$= 5 \times 4 \times 5 \times 4 = 25 \times 16 = 400$

AP EAMCET-05.10.2021

Permutation and Combination

119218 The remainder obtained when $(1!)^{2} + (2!)^{2} +$ $(3!)^{2} + \dots + (100!)^{2}$ is divided by $10^{2}$ is

1 27

2 28

3 17

4 14

Explanation:

C According to question,
$\frac{(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2} + (5!)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
$= \frac{1 + (2)^{2} + (6)^{2} + (24)^{2} + (120)^{2} + \dots \dots . (100!)^{2}}{10^{2}}$
Since, $(5!)^{2}, (6!)^{2}, (7!)^{2} \dots \dots (100!)^{2}$ each multiple of $(10)^{2}$
And for $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2}$
$∴ (1)^{2} + (2)^{2} + (6)^{2} + (24)^{2}$
$= 1 + 4 + 36 + 576$
$= 617$ When, 617 is divided by 100 its remainder is 17 .

COMEDK-2016

Permutation and Combination

119228 The English alphabets have 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed from the alphabet?

1

2100 \times 2

!

2

210 \times 2

!

3

210 \times 4

!

4

2100 \times 4

!

Explanation:

D 2 Vowels can be chosen in $^{5} C_{2}$ ways
And, 2 consonant can be chosen in $^{21} C_{2}$ ways.
4 letter can be arranged in 4 ways
$=^{5} C_{2} \times^{21} C_{2} \times 4$ !
$= \frac{5!}{2! (5 - 2)!} \times \frac{21!}{2! (21 - 2)!} \times 4$ !
$= \frac{5!}{2! 3!} \times \frac{21!}{2! 19!} \times 4$ !
$= \frac{5 \times 4 \times 3!}{2! 3!} \times \frac{21 \times 20 \times 19!}{2! \times 19!} \times 4$ !
$= 10 \times 210 \times 4! = 2100 \times 4$ !

AP EAMCET-07.07.2022

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