QBManager

Permutation and Combination

118976 In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls?

1

^{8} C_{5} \times^{10} C_{4}

2

^{10} C_{5} \times^{8} C_{4}

3

^{18} C_{9}

4

^{10} C_{4} \times^{8} C_{5}

Explanation:

B Number of ways in which $5 red balls can be$
$choosen among 10 red balls =^{10} C_{5}$
$And no. of ways in which 4 white balls can be choosen$
$among 8 white balls =^{8} C_{4}$
$∴ Total no. of ways =^{10} C_{5} \times^{8} C_{4}$

SRM JEEE-2011

Permutation and Combination

118977 The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of man and woman is

1 1240

2 1840

3 1820

4 2005

Explanation:

A Number of ways of choosing first couple
$consisting of 1 man and 1 woman =^{15} C_{1} \times^{15} C_{1} = (15)^{2}$
$No. of ways of choosing second couple consisting of 1$
$man and 1 woman = 14 C_{1} \times 14 C_{1}$
$= (14)^{2}$
$\sum (15)^{2} = \frac{15 \times (15 + 1) (30 + 1)}{6}$
$\sum (15)^{2} = \frac{15 \times 16 \times 31}{6} = 1240$
Similarly, 15 couples are formed
The total no. of ways 15 men and 15 women can be chosen to form a couple $= (15)^{2} + (14)^{2} + \dots \dots . . + (1)^{2}$
Now, $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

SRM JEEE-2014

Permutation and Combination

118978 The number of arrangements of the letters of the word BANANA is which the two ' $N$ 's do not appear adjacently is

1 40

2 60

3 80

4 100

Explanation:

A :
$∵$ Total words in BANANA $= 6$
Repeated words $= A, N$
So, the total number of permutation $= \frac{6!}{3! 2!}$
$= \frac{6 \times 5 \times 4}{2}$
$= 60$
Now, taking two of the N's together NNB, NAA
Taking NNN as one block $= \frac{5!}{3!}$
$= 5 \times 4 = 20$
Hence, the required permutation $= \frac{6!}{3! \times 2!} - \frac{5!}{3!}$
$= 60 - 20 = 40$

SRMJEE-2011]

Permutation and Combination

118979 The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

1 120

2 88

3 89

4 90

Explanation:

C Total words in TOUGH is 5.
So, the total rank $= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (24 \times 3) + (6 \times 2) + (4) + 1$
$= 72 + 12 + 4 + 1$
$= 89$

BITSAT-2009

Permutation and Combination

118980 How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

1 16

2 36

3 60

4 180

Explanation:

C Given, number $= 223355888$
Total digit is 9 .
Here repeated digit is $2, 3, 5, 8$
Four digit $3, 3, 5, 5$ , can be arranged at four places in
$= \frac{4!}{2! 2!}$
$= 6$
$2, 2, 8, 8, 8$ can be arranged at the five odd places in-
$= \frac{5!}{3! 2!}$
$= \frac{5 \times 4}{2} = 10$ So, the number of possible arrangement $= 6 \times 10 = 60$

BITSAT-2017

Permutation and Combination

118976 In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls?

1

^{8} C_{5} \times^{10} C_{4}

2

^{10} C_{5} \times^{8} C_{4}

3

^{18} C_{9}

4

^{10} C_{4} \times^{8} C_{5}

Explanation:

B Number of ways in which $5 red balls can be$
$choosen among 10 red balls =^{10} C_{5}$
$And no. of ways in which 4 white balls can be choosen$
$among 8 white balls =^{8} C_{4}$
$∴ Total no. of ways =^{10} C_{5} \times^{8} C_{4}$

SRM JEEE-2011

Permutation and Combination

118977 The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of man and woman is

1 1240

2 1840

3 1820

4 2005

Explanation:

A Number of ways of choosing first couple
$consisting of 1 man and 1 woman =^{15} C_{1} \times^{15} C_{1} = (15)^{2}$
$No. of ways of choosing second couple consisting of 1$
$man and 1 woman = 14 C_{1} \times 14 C_{1}$
$= (14)^{2}$
$\sum (15)^{2} = \frac{15 \times (15 + 1) (30 + 1)}{6}$
$\sum (15)^{2} = \frac{15 \times 16 \times 31}{6} = 1240$
Similarly, 15 couples are formed
The total no. of ways 15 men and 15 women can be chosen to form a couple $= (15)^{2} + (14)^{2} + \dots \dots . . + (1)^{2}$
Now, $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

SRM JEEE-2014

Permutation and Combination

118978 The number of arrangements of the letters of the word BANANA is which the two ' $N$ 's do not appear adjacently is

1 40

2 60

3 80

4 100

Explanation:

A :
$∵$ Total words in BANANA $= 6$
Repeated words $= A, N$
So, the total number of permutation $= \frac{6!}{3! 2!}$
$= \frac{6 \times 5 \times 4}{2}$
$= 60$
Now, taking two of the N's together NNB, NAA
Taking NNN as one block $= \frac{5!}{3!}$
$= 5 \times 4 = 20$
Hence, the required permutation $= \frac{6!}{3! \times 2!} - \frac{5!}{3!}$
$= 60 - 20 = 40$

SRMJEE-2011]

Permutation and Combination

118979 The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

1 120

2 88

3 89

4 90

Explanation:

C Total words in TOUGH is 5.
So, the total rank $= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (24 \times 3) + (6 \times 2) + (4) + 1$
$= 72 + 12 + 4 + 1$
$= 89$

BITSAT-2009

Permutation and Combination

118980 How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

1 16

2 36

3 60

4 180

Explanation:

C Given, number $= 223355888$
Total digit is 9 .
Here repeated digit is $2, 3, 5, 8$
Four digit $3, 3, 5, 5$ , can be arranged at four places in
$= \frac{4!}{2! 2!}$
$= 6$
$2, 2, 8, 8, 8$ can be arranged at the five odd places in-
$= \frac{5!}{3! 2!}$
$= \frac{5 \times 4}{2} = 10$ So, the number of possible arrangement $= 6 \times 10 = 60$

BITSAT-2017

Permutation and Combination

118976 In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls?

1

^{8} C_{5} \times^{10} C_{4}

2

^{10} C_{5} \times^{8} C_{4}

3

^{18} C_{9}

4

^{10} C_{4} \times^{8} C_{5}

Explanation:

B Number of ways in which $5 red balls can be$
$choosen among 10 red balls =^{10} C_{5}$
$And no. of ways in which 4 white balls can be choosen$
$among 8 white balls =^{8} C_{4}$
$∴ Total no. of ways =^{10} C_{5} \times^{8} C_{4}$

SRM JEEE-2011

Permutation and Combination

118977 The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of man and woman is

1 1240

2 1840

3 1820

4 2005

Explanation:

A Number of ways of choosing first couple
$consisting of 1 man and 1 woman =^{15} C_{1} \times^{15} C_{1} = (15)^{2}$
$No. of ways of choosing second couple consisting of 1$
$man and 1 woman = 14 C_{1} \times 14 C_{1}$
$= (14)^{2}$
$\sum (15)^{2} = \frac{15 \times (15 + 1) (30 + 1)}{6}$
$\sum (15)^{2} = \frac{15 \times 16 \times 31}{6} = 1240$
Similarly, 15 couples are formed
The total no. of ways 15 men and 15 women can be chosen to form a couple $= (15)^{2} + (14)^{2} + \dots \dots . . + (1)^{2}$
Now, $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

SRM JEEE-2014

Permutation and Combination

118978 The number of arrangements of the letters of the word BANANA is which the two ' $N$ 's do not appear adjacently is

1 40

2 60

3 80

4 100

Explanation:

A :
$∵$ Total words in BANANA $= 6$
Repeated words $= A, N$
So, the total number of permutation $= \frac{6!}{3! 2!}$
$= \frac{6 \times 5 \times 4}{2}$
$= 60$
Now, taking two of the N's together NNB, NAA
Taking NNN as one block $= \frac{5!}{3!}$
$= 5 \times 4 = 20$
Hence, the required permutation $= \frac{6!}{3! \times 2!} - \frac{5!}{3!}$
$= 60 - 20 = 40$

SRMJEE-2011]

Permutation and Combination

118979 The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

1 120

2 88

3 89

4 90

Explanation:

C Total words in TOUGH is 5.
So, the total rank $= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (24 \times 3) + (6 \times 2) + (4) + 1$
$= 72 + 12 + 4 + 1$
$= 89$

BITSAT-2009

Permutation and Combination

118980 How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

1 16

2 36

3 60

4 180

Explanation:

C Given, number $= 223355888$
Total digit is 9 .
Here repeated digit is $2, 3, 5, 8$
Four digit $3, 3, 5, 5$ , can be arranged at four places in
$= \frac{4!}{2! 2!}$
$= 6$
$2, 2, 8, 8, 8$ can be arranged at the five odd places in-
$= \frac{5!}{3! 2!}$
$= \frac{5 \times 4}{2} = 10$ So, the number of possible arrangement $= 6 \times 10 = 60$

BITSAT-2017

Permutation and Combination

118976 In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls?

1

^{8} C_{5} \times^{10} C_{4}

2

^{10} C_{5} \times^{8} C_{4}

3

^{18} C_{9}

4

^{10} C_{4} \times^{8} C_{5}

Explanation:

B Number of ways in which $5 red balls can be$
$choosen among 10 red balls =^{10} C_{5}$
$And no. of ways in which 4 white balls can be choosen$
$among 8 white balls =^{8} C_{4}$
$∴ Total no. of ways =^{10} C_{5} \times^{8} C_{4}$

SRM JEEE-2011

Permutation and Combination

118977 The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of man and woman is

1 1240

2 1840

3 1820

4 2005

Explanation:

A Number of ways of choosing first couple
$consisting of 1 man and 1 woman =^{15} C_{1} \times^{15} C_{1} = (15)^{2}$
$No. of ways of choosing second couple consisting of 1$
$man and 1 woman = 14 C_{1} \times 14 C_{1}$
$= (14)^{2}$
$\sum (15)^{2} = \frac{15 \times (15 + 1) (30 + 1)}{6}$
$\sum (15)^{2} = \frac{15 \times 16 \times 31}{6} = 1240$
Similarly, 15 couples are formed
The total no. of ways 15 men and 15 women can be chosen to form a couple $= (15)^{2} + (14)^{2} + \dots \dots . . + (1)^{2}$
Now, $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

SRM JEEE-2014

Permutation and Combination

118978 The number of arrangements of the letters of the word BANANA is which the two ' $N$ 's do not appear adjacently is

1 40

2 60

3 80

4 100

Explanation:

A :
$∵$ Total words in BANANA $= 6$
Repeated words $= A, N$
So, the total number of permutation $= \frac{6!}{3! 2!}$
$= \frac{6 \times 5 \times 4}{2}$
$= 60$
Now, taking two of the N's together NNB, NAA
Taking NNN as one block $= \frac{5!}{3!}$
$= 5 \times 4 = 20$
Hence, the required permutation $= \frac{6!}{3! \times 2!} - \frac{5!}{3!}$
$= 60 - 20 = 40$

SRMJEE-2011]

Permutation and Combination

118979 The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

1 120

2 88

3 89

4 90

Explanation:

C Total words in TOUGH is 5.
So, the total rank $= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (24 \times 3) + (6 \times 2) + (4) + 1$
$= 72 + 12 + 4 + 1$
$= 89$

BITSAT-2009

Permutation and Combination

118980 How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

1 16

2 36

3 60

4 180

Explanation:

C Given, number $= 223355888$
Total digit is 9 .
Here repeated digit is $2, 3, 5, 8$
Four digit $3, 3, 5, 5$ , can be arranged at four places in
$= \frac{4!}{2! 2!}$
$= 6$
$2, 2, 8, 8, 8$ can be arranged at the five odd places in-
$= \frac{5!}{3! 2!}$
$= \frac{5 \times 4}{2} = 10$ So, the number of possible arrangement $= 6 \times 10 = 60$

BITSAT-2017

Permutation and Combination

118976 In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls?

1

^{8} C_{5} \times^{10} C_{4}

2

^{10} C_{5} \times^{8} C_{4}

3

^{18} C_{9}

4

^{10} C_{4} \times^{8} C_{5}

Explanation:

B Number of ways in which $5 red balls can be$
$choosen among 10 red balls =^{10} C_{5}$
$And no. of ways in which 4 white balls can be choosen$
$among 8 white balls =^{8} C_{4}$
$∴ Total no. of ways =^{10} C_{5} \times^{8} C_{4}$

SRM JEEE-2011

Permutation and Combination

118977 The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of man and woman is

1 1240

2 1840

3 1820

4 2005

Explanation:

A Number of ways of choosing first couple
$consisting of 1 man and 1 woman =^{15} C_{1} \times^{15} C_{1} = (15)^{2}$
$No. of ways of choosing second couple consisting of 1$
$man and 1 woman = 14 C_{1} \times 14 C_{1}$
$= (14)^{2}$
$\sum (15)^{2} = \frac{15 \times (15 + 1) (30 + 1)}{6}$
$\sum (15)^{2} = \frac{15 \times 16 \times 31}{6} = 1240$
Similarly, 15 couples are formed
The total no. of ways 15 men and 15 women can be chosen to form a couple $= (15)^{2} + (14)^{2} + \dots \dots . . + (1)^{2}$
Now, $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

SRM JEEE-2014

Permutation and Combination

118978 The number of arrangements of the letters of the word BANANA is which the two ' $N$ 's do not appear adjacently is

1 40

2 60

3 80

4 100

Explanation:

A :
$∵$ Total words in BANANA $= 6$
Repeated words $= A, N$
So, the total number of permutation $= \frac{6!}{3! 2!}$
$= \frac{6 \times 5 \times 4}{2}$
$= 60$
Now, taking two of the N's together NNB, NAA
Taking NNN as one block $= \frac{5!}{3!}$
$= 5 \times 4 = 20$
Hence, the required permutation $= \frac{6!}{3! \times 2!} - \frac{5!}{3!}$
$= 60 - 20 = 40$

SRMJEE-2011]

Permutation and Combination

118979 The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

1 120

2 88

3 89

4 90

Explanation:

C Total words in TOUGH is 5.
So, the total rank $= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (4! \times 3) + (3! \times 2) + (2! \times 2) + 1$
$= (24 \times 3) + (6 \times 2) + (4) + 1$
$= 72 + 12 + 4 + 1$
$= 89$

BITSAT-2009

Permutation and Combination

118980 How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

1 16

2 36

3 60

4 180

Explanation:

C Given, number $= 223355888$
Total digit is 9 .
Here repeated digit is $2, 3, 5, 8$
Four digit $3, 3, 5, 5$ , can be arranged at four places in
$= \frac{4!}{2! 2!}$
$= 6$
$2, 2, 8, 8, 8$ can be arranged at the five odd places in-
$= \frac{5!}{3! 2!}$
$= \frac{5 \times 4}{2} = 10$ So, the number of possible arrangement $= 6 \times 10 = 60$

BITSAT-2017