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51. A college has 10 basketball players. A 5 member's team and a captain will be selected out of these 10 players. How many different selections can be made?
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210 × 6 = 1260
Alternatively
We can select the 5 member team out of the 10 in 10C5 ways = 252 ways.
The captain can be selected from amongst the remaining 5 players in 5 ways.
Therefore, total ways the selection of 5 players and a captain can be made,
= 252 × 5
= 1260
Solution:
A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210 × 6 = 1260
Alternatively
We can select the 5 member team out of the 10 in 10C5 ways = 252 ways.
The captain can be selected from amongst the remaining 5 players in 5 ways.
Therefore, total ways the selection of 5 players and a captain can be made,
= 252 × 5
= 1260
52. How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?
The first three letters can be either consonants or vowels. So, each of them have 26 options.
Note that the question asks you to find out the number of distinct initials and not initials where the letters are distinct.
Hence, required answer
= 26 × 26 × 26 × 21
= 263 × 21
Solution:
The last of the four letter words should be a consonant. Therefore, there are 21 options.The first three letters can be either consonants or vowels. So, each of them have 26 options.
Note that the question asks you to find out the number of distinct initials and not initials where the letters are distinct.
Hence, required answer
= 26 × 26 × 26 × 21
= 263 × 21
53. What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift?
Gf who will get 2 gifts can be find out in 8C1 ways = 8 ways.
Now 2 gifts can be given to selected gf in 9C2 ways. And remaining 7 gifts can be given to remaining 7 gf in 7! ways.
So total no of ways= 8 × 9C2 × 7!
=
= 36 × 8 × 7!
= 36 × 8!
Solution:
One among 8 gfs will get 2 gifts and remaining 7 will get one. So total of 9 gifts will be distributed among 8 gfs.i.e; 11111112Gf who will get 2 gifts can be find out in 8C1 ways = 8 ways.
Now 2 gifts can be given to selected gf in 9C2 ways. And remaining 7 gifts can be given to remaining 7 gf in 7! ways.
So total no of ways= 8 × 9C2 × 7!
=
= 36 × 8 × 7!
= 36 × 8!
54. How many number of times will the digit 7 be written when listing the integers from 1 to 1000?
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1 × 9 × 9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3 × 81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 × 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
= 243 + 54 + 3
= 300
Solution:
7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9. 1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1 × 9 × 9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3 × 81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 × 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
= 243 + 54 + 3
= 300
55. In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements =
You can choose the 7 people to sit in the first table in 15C7 ways.
After selecting 7 people for the table that can seat 7 people, they can be seated in:
(7 - 1)! = 6!
The remaining 8 people can be made to sit around the second circular table in:(8 - 1)! = 7! Ways.
Hence, total number of ways: 15C7 × 6! × 7!
Solution:
'n' objects can be arranged around a circle in (n - 1)! ways.If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements =
You can choose the 7 people to sit in the first table in 15C7 ways.
After selecting 7 people for the table that can seat 7 people, they can be seated in:
(7 - 1)! = 6!
The remaining 8 people can be made to sit around the second circular table in:(8 - 1)! = 7! Ways.
Hence, total number of ways: 15C7 × 6! × 7!
56. In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?
The vowels occupy 3rd, 5th, 7th and 8th position in the word and the remaining 5 positions are occupied by consonants.
As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the before mentioned 4 places and the consonants can occupy 1st, 2nd, 4th, 6th and 9th positions.
The 4 vowels can be arranged in the 3rd, 5th, 7th and 8th position in 4! Ways.
Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th, 6th and 9th
Solution:
The word EDUCATION is a 9 letter word, with none of the letters repeating.The vowels occupy 3rd, 5th, 7th and 8th position in the word and the remaining 5 positions are occupied by consonants.
As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the before mentioned 4 places and the consonants can occupy 1st, 2nd, 4th, 6th and 9th positions.
The 4 vowels can be arranged in the 3rd, 5th, 7th and 8th position in 4! Ways.
Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th, 6th and 9th
57. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?
We may divide the 8 students as follows:
Case I:
5 students in the first car and 3 in the second.
Hence, 8 students are divided into groups of 5 and 3 in 8C3 = 56 ways.
Case II:
4 students in the first car and 4 in the second.
So, 8 students are divided into two groups of 4 and 4 in 8C4 = 78 ways.
Therefore, the total number of ways in which 8 students can travel is:
56 + 70 = 126
Solution:
There are 8 students and the maximum capacity of the cars together is 9.We may divide the 8 students as follows:
Case I:
5 students in the first car and 3 in the second.
Hence, 8 students are divided into groups of 5 and 3 in 8C3 = 56 ways.
Case II:
4 students in the first car and 4 in the second.
So, 8 students are divided into two groups of 4 and 4 in 8C4 = 78 ways.
Therefore, the total number of ways in which 8 students can travel is:
56 + 70 = 126
58. A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible ways of presenting these 10 cups to the expert.
Solution:
Since, there are 5 cups of each kind, prepared with milk or tea leaves added first, are identical hence, total number of different people ways of presenting the cups to the expert is,
59. A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 woman. The shortlist consists of 9 men and 6 women. In how many ways can this committee be formed?
= (6C1 × 9C6) + (6C2 × 9C5) + (6C3 × 9C4)
= 4914
Solution:
Three possibilities: (1W+6M), (2W+5M), (3W+4M)= (6C1 × 9C6) + (6C2 × 9C5) + (6C3 × 9C4)
= 4914
60. How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
Let the number of required alphabets in the language be 'n'.
Therefore, using 'n' alphabets we can form n × n × n = n3 distinct 3 digit initials.
NOTE:
Distinct initials are different from initials where the digits are different. For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This n3 different initials = 1 million.
i.e. n3 = 106 (1 million =106)
n3 = 1023
n = 102
n = 100
Hence, the language needs to have a minimum of 100 alphabets
Solution:
1 million distinct 3 digit initials are needed. Let the number of required alphabets in the language be 'n'.
Therefore, using 'n' alphabets we can form n × n × n = n3 distinct 3 digit initials.
NOTE:
Distinct initials are different from initials where the digits are different. For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This n3 different initials = 1 million.
i.e. n3 = 106 (1 million =106)
n3 = 1023
n = 102
n = 100
Hence, the language needs to have a minimum of 100 alphabets