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81. A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?
We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.
If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.
There is only one way for no wrong packages delivered to occur.
This is the same as everyone gets the right package.
The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
⇒ 1 × 1 × 1 = 1
Determine the total number of ways the three packages can be delivered.
⇒ 3 × 2 × 1 = 6
The number of ways at least one house gets the wrong package is:
⇒ 6 - 1 = 5
Therefore there are 5 ways for at least one house to get the wrong package.
Solution:
The possible outcomes that satisfy the condition of at least one house gets the wrong package are:One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.
If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.
There is only one way for no wrong packages delivered to occur.
This is the same as everyone gets the right package.
The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
⇒ 1 × 1 × 1 = 1
Determine the total number of ways the three packages can be delivered.
⇒ 3 × 2 × 1 = 6
The number of ways at least one house gets the wrong package is:
⇒ 6 - 1 = 5
Therefore there are 5 ways for at least one house to get the wrong package.
82. How many natural numbers less than a lakh can be formed with the digits 0,6 and 9?
The required numbers are from 1 to 99999
The numbers are five digit numbers.
Therefore, every place can be filled by 0, 6 and 9 in 3 ways.
Total number of ways = 3 × 3 × 3 × 3 × 3 = 35
But 00000 is also a number formed and has to be excluded.
Total number of numbers,
= 35 - 1
= 243 - 1
= 242
Solution:
The digits to be used are 0,6 and 9The required numbers are from 1 to 99999
The numbers are five digit numbers.
Therefore, every place can be filled by 0, 6 and 9 in 3 ways.
Total number of ways = 3 × 3 × 3 × 3 × 3 = 35
But 00000 is also a number formed and has to be excluded.
Total number of numbers,
= 35 - 1
= 243 - 1
= 242
83. There are 20 couples in a party. Every person greets every person except his or her spouse. People of the same sex shake hands and those of opposite sex greet each other with a Namaste (It means bringing one's own palms together and raising them to the chest level). What is the total number of handshakes and Namaste's in the party?
When a man meets a woman, there are two Namastes, whereas when a man meets a man (or a woman) there is only 1 handshake.
Number of handshakes,
= 2 × 20C2 (men and women ) = 2 × 190 = 380
For a number of Namastes, Every man does 19 Namastes (to the 20 women excluding his wife) and they respond in the same way.
Number of Namastes
= 2 × 20 × 19
= 760
Total number of Namastes and Handshake
= 760 + 380
= 1140
Solution:
There are 20 men and 20 women.When a man meets a woman, there are two Namastes, whereas when a man meets a man (or a woman) there is only 1 handshake.
Number of handshakes,
= 2 × 20C2 (men and women ) = 2 × 190 = 380
For a number of Namastes, Every man does 19 Namastes (to the 20 women excluding his wife) and they respond in the same way.
Number of Namastes
= 2 × 20 × 19
= 760
Total number of Namastes and Handshake
= 760 + 380
= 1140
84. In how many ways can a leap year have 53 Sundays?
So to have 53 Sundays one of these two days must be a Sunday.
This can occur in only 2 ways.
i.e. (Saturday and Sunday) or (Sunday and Monday).
Thus number of ways = 2
Solution:
In a leap year there are 366 days i.e. 52 weeks + 2 extra days.So to have 53 Sundays one of these two days must be a Sunday.
This can occur in only 2 ways.
i.e. (Saturday and Sunday) or (Sunday and Monday).
Thus number of ways = 2
85. In a certain laboratory, chemicals are identified by a colour-coding system. There are 20 different chemicals. Each one is coded with either a single colour or a unique two-colour pair. If the order of colours in the pairs does not matter, what is the minimum number of different colours needed to code all 20 chemicals with either a single colour or a unique pair of colours?
Therefore, total number of ways = n + nC2
Minimum number of different colour needed to code all 20 chemicals will be 6
= 6 + 6C2
=21
Solution:
Each one coded with either a single colour or unique two-colour pair.Therefore, total number of ways = n + nC2
Minimum number of different colour needed to code all 20 chemicals will be 6
= 6 + 6C2
=21
86. Six boxes are numbered 1, 2, 3, 4, 5 and 6. Each box must contain either a white ball or a black ball. At least one box must contain a black ball and boxes containing black balls must be consecutively numbered. find the total number of ways of placing the balls.
If there are 2 black balls, they can be placed in 5 ways (in 1,2 ; 2,3 ; 3,4 ; 4,5 and 5,6) and so on.
If there are 6 black balls, they can be placed in 1 way.
The total number of ways of placing the balls is
= 1 + 2 + 3 + 4 + 5 + 6
= 21
Solution:
If there is 1 black ball, it can be placed in 6 ways.If there are 2 black balls, they can be placed in 5 ways (in 1,2 ; 2,3 ; 3,4 ; 4,5 and 5,6) and so on.
If there are 6 black balls, they can be placed in 1 way.
The total number of ways of placing the balls is
= 1 + 2 + 3 + 4 + 5 + 6
= 21
87. In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together whereas 2 particular green toys are always together?
We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2 who are not to be together) is:
= 9! × 2! × 10C2 × 2!
= 18 × 10!
Solution:
Considering two green toys that are to be together as one unit.We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2 who are not to be together) is:
= 9! × 2! × 10C2 × 2!
= 18 × 10!
88. There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-3 are never together?
The total number of ways in which part-1 and part-3 are always together:
= 4! × 2!
= 48
Therefore, the total number of arrangements, in which they are not together is:
= 120 - 48
= 72
Solution:
The total number of ways in which 5 part can be arranged =5! =120.The total number of ways in which part-1 and part-3 are always together:
= 4! × 2!
= 48
Therefore, the total number of arrangements, in which they are not together is:
= 120 - 48
= 72
89. The number of ways which a mixed double tennis game can be arranged amongst 9 married couples if no husband and wife play in the same is:
We know that in a mixed double match there are two males and two females.
Step I: Two male members can be selected in 9C2 = 36 ways.
Step II: Having selected two male members, 2 female members can be selected in,
7C2 = 21 ways.
Step III: Two male and two female members can arranged in a particular game in 2 ways.
Total number of arrangements
= 36 × 21 × 2
= 1512 ways
Solution:
As per the question there are 9 married couples and no husband and wife should play in the same game:We know that in a mixed double match there are two males and two females.
Step I: Two male members can be selected in 9C2 = 36 ways.
Step II: Having selected two male members, 2 female members can be selected in,
7C2 = 21 ways.
Step III: Two male and two female members can arranged in a particular game in 2 ways.
Total number of arrangements
= 36 × 21 × 2
= 1512 ways
90. Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?
When 10 coins are tossed simultaneously, the total number of outcomes = 210
Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head.
Therefore, the remaining 9 coins can turn up either a head or a tail = 29
Solution:
When a coin is tossed once, there are two outcomes. It can turn up a head or a tail.When 10 coins are tossed simultaneously, the total number of outcomes = 210
Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head.
Therefore, the remaining 9 coins can turn up either a head or a tail = 29