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41. There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain three women and one man?
Let's sort out what we have and what we want.
Have: 5 women, 6 men.
Want: 3 women AND 1 man.
The word AND means multiply.
Woman and Men
Solution:
Since, no order to the committee is mentioned, a combination instead of a permutation is used.Let's sort out what we have and what we want.
Have: 5 women, 6 men.
Want: 3 women AND 1 man.
The word AND means multiply.
Woman and Men
42. Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct.
Coming to numbers part, there are 10 ways (any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10 ways to choose the second digit.
Hence, there are totally 10 × 10 = 100 ways.
Combined with letters there are,
26P2 × 100 = 65000 ways to choose vehicle numbers.
Solution:
Out of 26 alphabets two distinct letters can be chosen in 26P2 ways.Coming to numbers part, there are 10 ways (any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10 ways to choose the second digit.
Hence, there are totally 10 × 10 = 100 ways.
Combined with letters there are,
26P2 × 100 = 65000 ways to choose vehicle numbers.
43. In a railway compartment, there are 2 rows of seats facing each other with accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the rear while 3 others are indifferent. In how many ways can the 10 passengers be seated?
Total number of ways
= 5P4 × 5P3 × 3P3
= 43200
Solution:
The four person who wish to sit facing forward can be seated in: 5P4 ways and 3 who wish to sit facing towards the rear can be seated in: 5P3 ways and the remaining 3 can be seated in the remaining 3 seats in 3P3 ways.Total number of ways
= 5P4 × 5P3 × 3P3
= 43200
44. There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in:
5C3 ways = 10 ways.
Solution:
3 prize among 5 students can be distributed in,5C3 ways = 10 ways.
45. A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can, without taking the same pair of children together more than once. How many times does the teacher go to the zoo?
Therefore, the teacher goes to the zoo 15 times.
Solution:
Two students can be selected from 6 in 6C2 =15 ways.Therefore, the teacher goes to the zoo 15 times.
46. A family consist of a grandfather, 5 sons and daughter and 8 grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each end and the grandfather refuses to have a grandchild on either side of him. The number of ways in which the family can be made to sit is:
= 1 grandfather + 5 sons and daughters + 8 grandchildren
= 14
The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways.
The grandfather can occupy a seat in (5 - 1) = 4 ways (4 gaps between 5 sons and daughter).
And, the remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter).
Hence total number of required ways,
= 8! × 480
Solution:
Total no. of seats,= 1 grandfather + 5 sons and daughters + 8 grandchildren
= 14
The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways.
The grandfather can occupy a seat in (5 - 1) = 4 ways (4 gaps between 5 sons and daughter).
And, the remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter).
Hence total number of required ways,
= 8! × 480
47. If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?
The first 24 of these words will start with A.
Then, the 25th word will start will CA _ _ _ .
The remaining 3 letters can be rearranged in 3! = 6 Ways.
i.e. 6 words exist that start with CA.
The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.
Therefore, the rank of CHASM will be 24 + 6 + 2 = 32
Solution:
The 5 letter word can be rearranged in 5! = 120 Ways without any of the letters repeating.The first 24 of these words will start with A.
Then, the 25th word will start will CA _ _ _ .
The remaining 3 letters can be rearranged in 3! = 6 Ways.
i.e. 6 words exist that start with CA.
The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.
Therefore, the rank of CHASM will be 24 + 6 + 2 = 32
48. In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways:
Case i. 2, 2, 1
Case ii. 3, 1, 1
Case i:
Number of ways of achieving the first option 2, 2, 1
Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way.
However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.
Therefore, total number of ways of achieving the 2, 2, 1 option is:
Case ii: Number of ways of achieving the second option 3, 1, 1
Three toys out of the 5 can be selected in ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3, 1, 1 option is = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case i + number of ways of achieving Case ii
= 15 + 10
= 25 ways.
Solution:
The toys are different; The boxes are identical.If none of the boxes is to remain empty, then we can pack the toys in one of the following ways:
Case i. 2, 2, 1
Case ii. 3, 1, 1
Case i:
Number of ways of achieving the first option 2, 2, 1
Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way.
However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.
Therefore, total number of ways of achieving the 2, 2, 1 option is:
Case ii: Number of ways of achieving the second option 3, 1, 1
Three toys out of the 5 can be selected in ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3, 1, 1 option is = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case i + number of ways of achieving Case ii
= 15 + 10
= 25 ways.
49. What is the value of 1 × 1! + 2 × 2! + 3 × 3! + . . . . . . . . n × n!
where n! means n factorial or n(n-1) (n-2) . . . . . . . . 1
2 × 2! = (3 - 1) × 2! = 3 × 2! - 2! = 3! - 2!
3 × 3! = (4 - 1) × 3! = 4 × 3! - 3! = 4! - 3!
..
..
..
n × n! = (n + 1 - 1) × n! = (n + 1) (n!) - n! = (n + 1)! - n!
Summing up all these terms, we get (n + 1)! - 1!
where n! means n factorial or n(n-1) (n-2) . . . . . . . . 1
Solution:
1 × 1! = (2 - 1) × 81! = 2 × 1! - 1 × 1! = 2! - 1!2 × 2! = (3 - 1) × 2! = 3 × 2! - 2! = 3! - 2!
3 × 3! = (4 - 1) × 3! = 4 × 3! - 3! = 4! - 3!
..
..
..
n × n! = (n + 1 - 1) × n! = (n + 1) (n!) - n! = (n + 1)! - n!
Summing up all these terms, we get (n + 1)! - 1!
50. When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.
The number of outcomes in which 0 coins turn heads is,
6C0 = 1 outcome.
The number of outcomes in which 1 coin turns head is,
6C1 = 6 outcomes.
The number of outcomes in which 2 coins turn heads is,
6C2 = 15 outcomes.
The number of outcomes in which 3 coins turn heads is,
6C3 = 20 outcomes.
Therefore, total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.
Solution:
The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.
The number of outcomes in which 0 coins turn heads is,
6C0 = 1 outcome.
The number of outcomes in which 1 coin turns head is,
6C1 = 6 outcomes.
The number of outcomes in which 2 coins turn heads is,
6C2 = 15 outcomes.
The number of outcomes in which 3 coins turn heads is,
6C3 = 20 outcomes.
Therefore, total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.