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If there are $$M$$ ways from $$P$$ to $$Q$$ and $$N$$ ways to go from $$Q$$ to $$R$$, then the total number of ways of going from $$P$$ to $$R$$ is $$M \times N$$
Addition principle: If event A can occur in m ways and event B in n ways and they are mutually exclusive, total = m+n
The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is
Correct Answer: 47
The possible arrangements are of the form
35 _ Can be chosen in 6 ways.
35 _ _ We can choose 2 out of the remaining 6 in $$^6C_2=15$$ways. We remove 1 case where 7 and 8 are together to get 14 ways.
35 _ _ _We can choose 3 out of the remaining 6 in $$^6C_3=20$$ways. We remove 4 cases where 7 and 8 are together to get 16 ways.
35 _ _ _ _We can choose 4 out of the remaining 6 in $$^6C_4=15$$ways. We remove 6 case where 7 and 8 are together to get 9 ways.
35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.
Thus, total number of cases = 6+14+16+9+2 = 47.
Alternatively,
The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.
Considering the cases, only 7 is selected.
We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.
Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have
Each of the numbers can either be selected or not selected and we have 4 numbers :
Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.
SImilarly, including only 8, we have 16 more possibilities.
Cases including neither 7 nor 8.
We must have 3 and 5 in the group but there must be no 7 and 8 in the group.
Hence we have 3 5 _ _ _ _.
For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.
= 16 possibilities
But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.
Hence 16 - 1 = 15 cases.
Total = 16+15+16 = 47 possibilities
How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?
Correct Answer: 70
Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.
Then (100c+10b+a)-(100a+10b+c)=198
99c-99a=198
c-a = 2.
Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.
Thus, there can be 7 cases.
B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be $$7\times\ 10=70$$.
The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is
Correct Answer: 1000
This question is an application of the product rule in probability and combinatorics.
In the product rule, if two events A and B can occur in x and y ways, and for an event E, both events A and B need to take place, the number of ways that E can occur is xy. This can be expanded to 3 or more events as well.
Event 1: Distribution of balloons
Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.
So we are left with 15 - 4 x 3 = 15 - 12 = 3 balloons and 3 children.
Now we need to distribute 3 identical balloons to 3 children.
This can be done in $$^{n+r-1}C_{r-1}$$ ways, where n = 3 and r = 3.
So, number of ways = $$^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$$
Event 2: Distribution of pencils
Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with 6 - 3 = 3 pencils.
We now need to distribute 3 identical pencils to 3 children.
This can be done in $$^{n+r-1}C_{r-1}$$ ways, where n = 3 and r = 3.
So, number of ways = $$^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$$
Event 3: Distribution of erasers
We need to distribute 3 identical erasers to 3 children.
This can be done in $$^{n+r-1}C_{r-1}$$ ways, where n = 3 and r = 3.
So, number of ways = $$^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$$
Applying the product rule, we get the total number of ways = 10 x 10 x 10 = 1000.
A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is
Correct Answer: 50
The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.
The different possibilities include :
Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :
$$\frac{4!}{3!}$$ = 4.
Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :
$$\frac{4!}{2!\cdot2!}=\ 6$$
Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :
$$\frac{4!}{3!}$$ = 4
Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :
$$\frac{4!}{2!}=\ 12$$
Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :
$$\frac{4!}{2!}=\ 12$$
Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :
$$\frac{4!}{2!}=\ 12$$
A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.
Alternatively
We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once
So the possible cases :
Now we get $$\frac{4!}{2!}\times\ 3$$= 36 ( When one digit is used twice and the remaining two once )
$$\frac{4!}{3!}\times\ 2$$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$$\frac{4!}{2!\times\ 2!}=\ 6$$( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50
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