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91. A book-shelf can accommodate 6 books from left to right. If 10 identical books on each of the languages A,B,C and D are available, In how many ways can the book shelf be filled such that book on the same languages are not put adjacently.
The subsequent places can be filled in 3 ways each.
Hence, the number of ways
= 4 × 3 × 3 × 3 × 3 × 3
= 4 × 35
Solution:
First place can be filled in 4 ways.The subsequent places can be filled in 3 ways each.
Hence, the number of ways
= 4 × 3 × 3 × 3 × 3 × 3
= 4 × 35
92. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together.
One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged.
The number of possible rearrangements is 4!
The group of 3 vowels contains two a s and one u
The 3 vowels can rearrange amongst themselves in ways as the vowel a appears twice.
Hence, the total number of rearrangements in which the vowels appear together are:
Solution:
ABACUS is a 6 letter word with 3 of the letters being vowels.If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together.
One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged.
The number of possible rearrangements is 4!
The group of 3 vowels contains two a s and one u
The 3 vowels can rearrange amongst themselves in ways as the vowel a appears twice.
Hence, the total number of rearrangements in which the vowels appear together are:
93. If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number:
If it started with c then the remaining 5 positions can be filled in 5! Ways.
Similarly if it started with H, I, N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A, C, H, I, N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5 × 5! = 600 Ways.
i.e. SACHIN is the 601th word.
Solution:
If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.If it started with c then the remaining 5 positions can be filled in 5! Ways.
Similarly if it started with H, I, N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A, C, H, I, N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5 × 5! = 600 Ways.
i.e. SACHIN is the 601th word.
94. In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?
However, the problem introduces a constraint that no man stands in a position that is ahead of his wife.
For any 2 given positions out of the 6 occupied by a man and his wife, the pair cannot rearrange amongst themselves in 2! Ways as the wife has to be in a position ahead of the man. Only one of the 2! arrangements is allowed.
As there are 3 couples in the group, the total number of ways gets reduced by a factor of
(2! × 2! × 2!)
Hence, the total number of ways,
=
Solution:
6 people can be made to stand in a line in 6! Ways.However, the problem introduces a constraint that no man stands in a position that is ahead of his wife.
For any 2 given positions out of the 6 occupied by a man and his wife, the pair cannot rearrange amongst themselves in 2! Ways as the wife has to be in a position ahead of the man. Only one of the 2! arrangements is allowed.
As there are 3 couples in the group, the total number of ways gets reduced by a factor of
(2! × 2! × 2!)
Hence, the total number of ways,
=
95. In a cricket match if a batsman score 0, 1, 2, 3, 4 or 6 runs of a ball, then find the number or different sequences in which he can score exactly 30 runs of an over. Assume that an over consists of only 6 balls and there were no extra and no run outs.
Case B: Four 6 and one '2' and one '4' = = 30
Case C: Three 6 and three '4' = = 20
Case D: Four 6 and two '3' = = 15
Total number of different sequences = 71
Solution:
Case A: Five 6 and one 'zero' = = 6Case B: Four 6 and one '2' and one '4' = = 30
Case C: Three 6 and three '4' = = 20
Case D: Four 6 and two '3' = = 15
Total number of different sequences = 71
96. In how many ways can the letters of the word MANAGEMENT be rearranged so that the two As do not appear together?
Normally, any 10 letter word can be rearranged in 10! ways.
However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each.
Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to:
The problem requires us to find out the number of outcomes in which the two As do not appear together.
The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter.
Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in:
ways.
Therefore, the required answer in which the two As do not appear next to each other
Solution:
The word MANAGEMENT is a 10 letter word.Normally, any 10 letter word can be rearranged in 10! ways.
However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each.
Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to:
The problem requires us to find out the number of outcomes in which the two As do not appear together.
The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter.
Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in:
ways.
Therefore, the required answer in which the two As do not appear next to each other
97. How many numbers are there between 100 and 1000 such that at least one of their digits is 6?
Numbers between 100 and 1000 which do not have digit 6 in any place,
= 8 × 9 × 9
= 648
Unit digit could take any value of the 9 values (0 to 9, except 6)
Tens Digit could take any value of the 9 values (0 to 9, except 6)
Hundreds digit could take any value of the 8 values (1 to 9, except 6)
numbers between 100 and 1000 which have at least one digit as 6,
= 900 - 648
= 252
Solution:
numbers between 100 and 1000 = 900Numbers between 100 and 1000 which do not have digit 6 in any place,
= 8 × 9 × 9
= 648
Unit digit could take any value of the 9 values (0 to 9, except 6)
Tens Digit could take any value of the 9 values (0 to 9, except 6)
Hundreds digit could take any value of the 8 values (1 to 9, except 6)
numbers between 100 and 1000 which have at least one digit as 6,
= 900 - 648
= 252
98. A student is required to answer 6 out of 10 questions divided into two groups each containing 5 questions. He is not permitted to attempt more than 4 from each group. In how many ways can he make the choice?
Number of ways of attempting more than 4 from a group,
= 2 × 5C5 × 5C1
= 10
Required number of ways
= 210 - 10
= 200
Solution:
Number of ways of choosing 6 from 10 = 10C6 = 210Number of ways of attempting more than 4 from a group,
= 2 × 5C5 × 5C1
= 10
Required number of ways
= 210 - 10
= 200
99. While packing for a business trip Mr. Debashis has packed 3 pairs of shoes, 4 pants, 3 half-pants,6 shirts, 3 sweater and 2 jackets. The outfit is defined as consisting of a pair of shoes, a choice of "lower wear" (either a pant or a half-pant), a choice of "upper wear" (it could be a shirt or a sweater or both) and finally he may or may not choose to wear a jacket. How many different outfits are possible?
= 3C1
= 3 ways
Number of ways lower wear can be selected = (3 + 4) = 7 ways.
Number of ways upper wear can be selected,= 3 + 6 + (3 × 6)
= 27 ways
Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1st jacket, 2nd jacket}.
Total number of different outfits
= 3 × 7 × 27 × 3
= 1701 ways
Solution:
Number of ways a pair of shoes can be selected,= 3C1
= 3 ways
Number of ways lower wear can be selected = (3 + 4) = 7 ways.
Number of ways upper wear can be selected,= 3 + 6 + (3 × 6)
= 27 ways
Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1st jacket, 2nd jacket}.
Total number of different outfits
= 3 × 7 × 27 × 3
= 1701 ways
100. How many positive integers 'n' can be form using the digits 3, 4, 4, 5, 6, 6, 7 if we want 'n' to exceed 60,00,000?
If the first digit is 6, the other digits can be arranged in = 360 ways.
If the first digit is 7, the other digits can be arranged in = 180 ways.
Thus required possibilities for n,
= 360 + 180
= 540 ways
Solution:
As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways.If the first digit is 6, the other digits can be arranged in = 360 ways.
If the first digit is 7, the other digits can be arranged in = 180 ways.
Thus required possibilities for n,
= 360 + 180
= 540 ways
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