Permutations and Combinations
The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is
If for some $m, n ;{ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3$ and ${ }^{n-1} P_3:{ }^n P_4=1: 8$, then ${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$ is equal to
Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to
Let $\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$ and $\beta=\frac{(5 !) !}{(5 !)^{4 !}}$. Then :
Explanation:
Since the greatest digit is 7, the possible digits for the pin code are $0, 1, 2, 3, 4, 5, 6,$ and $7$.
1. Let's denote the sum of the first two digits and the sum of the last two digits as $\lambda$. Since the largest digit is 7, we can analyze different possible values for $\lambda$.
We will analyze different possible values for $\lambda$ and the corresponding possible pairs of digits:
- $\lambda = 7$: The pairs of digits that have a sum of 7 are $(0,7),(7,0),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. Since 7 must be present, we consider the pairs with 7: $(0,7),(7,0)$.
When the pair $(0,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $07xy$ and $07yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.
When the pair $(7,0)$ is chosen, the other two digits can also be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $70xy$ and $70yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.
So, there are 24 possible pin codes for $\lambda = 7$.
- $\lambda = 8$: The pairs of digits that have a sum of 8 are $(1,7),(7,1),(2,6),(6,2),(3,5),(5,3)$. Since 7 must be present, we consider the pairs with 7: $(1,7),(7,1)$.
When the pair $(1,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $17xy$ and $17yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
When the pair $(7,1)$ is chosen, the other two digits can also be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $71xy$ and $71yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
So, there are 16 possible pin codes for $\lambda = 8$.
- $\lambda = 9$: The pairs of digits that have a sum of 9 are $(2,7),(7,2),(3,6),(6,3),(4,5),(5,4)$. Since 7 must be present, we consider the pairs with 7: $(2,7),(7,2)$.
When the pair $(2,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $27xy$ and $27yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
When the pair $(7,2)$ is chosen, the other two digits can also be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $72xy$ and $72yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
So, there are 16 possible pin codes for $\lambda = 9$.
- $\lambda = 10$: The pairs of digits that have a sum of 10 are $(3,7),(7,3),(4,6),(6,4)$. Since 7 must be present, we consider the pairs with 7: $(3,7),(7,3)$.
When the pair $(3,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $37xy$ and $37yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
When the pair $(7,3)$ is chosen, the other two digits can also be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $73xy$ and $73yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
So, there are 8 possible pin codes for $\lambda = 10$.
- $\lambda = 11$: The pairs of digits that have a sum of 11 are $(4,7),(7,4),(5,6),(6,5)$. Since 7 must be present, we consider the pairs with 7: $(4,7),(7,4)$.
When the pair $(4,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $47xy$ and $47yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
When the pair $(7,4)$ is chosen, the other two digits can also be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $74xy$ and $74yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
So, there are 8 possible pin codes for $\lambda = 11$.
- $\lambda = 12$: The only pair of digits that have a sum of 12 is $(5,7)$. Since 7 must be present, this case is not possible.
- $\lambda = 13$: The only pair of digits that have a sum of 13 is $(6,7)$. Since 7 must be present, this case is not possible.
- $\lambda = 14$: The only pair of digits that have a sum of 14 is $(7,7)$. Since all the digits must be different, this case is not possible.
Adding up all the possible pin codes for each value of $\lambda$, we get:
$16 + 24 + 24 + 16 + 8 = 72$
Therefore, the maximum number of trials necessary to obtain the correct code is 72.
Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits $1,2,3,4,5$ with repetition, is _________.
Explanation:
$ \begin{aligned} & (2,1,3),(2,3,4),(2,5,5),(2,2,5),(2,2,2) \\\\ & (4,1,1),(4,4,1),(4,4,4),(4,3,5) \\\\ & 2,1,3 \Rightarrow 312,132 \\\\ & 2,3,4 \Rightarrow 342,432,234,324 \\\\ & 2,5,5 \Rightarrow 552 \\\\ & 2,2,5 \Rightarrow 252,522 \\\\ & 2,2,2 \Rightarrow 222 \\\\ & 4,1,1 \Rightarrow 114 \\\\ & 4,4,1 \Rightarrow 414,144 \\\\ & 4,4,4 \Rightarrow 444 \\\\ & 4,3,5 \Rightarrow 354,534 \end{aligned} $
Total 16 numbers.
The number of seven digit positive integers formed using the digits $1,2,3$ and $4$ only and sum of the digits equal to $12$ is ___________.
Explanation:
Number of solutions
$ \begin{aligned} & =\text { Coefficient of } x^{12} \text { in }\left(x^1+x^2+x^3+x^4\right)^7 \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1+x+x^2+x^3\right)^7 \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1-x^4\right)^7(1-x)^{-7} \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right)(1-x)^{-7} \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right) \sum_{r=0}^{\infty}{ }^{7+r-1} C_r \cdot x^r \\\\ & ={ }^{11} C_5-7 \times{ }^7 C_1 \\\\ & =462-49=413 \end{aligned} $
Let the digits a, b, c be in A. P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?
Explanation:
The problem involves forming nine-digit numbers from three digits a, b, c which are in Arithmetic Progression (AP), used three times each, such that at least once, three consecutive digits are in AP.
We have the two possible sequences for the AP :
- a, b, c
- c, b, a
This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.
The next step is to choose the location of this sequence of three numbers within our nine-digit number.
Since there are nine places in our number and our sequence takes up three places, we have seven different starting points for our sequence : it can start at the first place, the second place, and so on, up to the seventh place.
Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7.
However, we also have to account for the fact that our sequence can be in one of two orders (a, b, c or c, b, a). So, we multiply the number of starting points by 2 to get $^7C_1 \times 2 = 14$ ways to arrange the sequence within our nine-digit number.
The remaining 6 digits (two 'a', two 'b', two 'c") can be arranged in $\frac{6!}{2!2!2!}$ ways.
Therefore, the total number of such nine-digit numbers is $^7C_1 \times 2 \times \frac{6!}{2!2!2!} = 1260$.
In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _________.
Explanation:
The formula for calculating the number of derangements (also known as subfactorials) is :
D(n) $ =n !\left[\frac{1}{0!}-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots . .+(-1)^n \frac{1}{n !}\right] $
Where n is the number of students, in this case, 5.
Using the formula, let's calculate the derangements for 5 students :
D(5) = $5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)$
D(5) = $120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
D(5) = $120 \left(0 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
D(5) = $120 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
D(5) = $120 \left(\frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120}\right)$
D(5) = $120 \left(\frac{44}{120}\right)$
D(5) = 44
So, there are 44 ways in which none of the students sits on the allotted seat.
The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to __________.
Explanation:
Total numbers when 1 is at unit digit $=\frac{3 !}{2 !}=3$
Total number when 2 is at unit digit $=3 !=6$
Total numbers when 3 is at unit digit $=\frac{3 !}{2 !}=3$
Sum of digits at unit place $=3 \times 1+6 \times 2+3 \times 3=24$
$\therefore$ Required sum $=24 \times 1000+24 \times 100+24 \times 10+24 \times 1$
$=24 \times 1111=26664$
The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ___________.
Explanation:
Total permutations $=7$!
Let $p=$ Number which containing string 153
$q=$ Number which containing string 2467
$ \begin{array}{ll} & \therefore n(p)=5! \times 1 \\\\ & \Rightarrow n(q)=4! \times 1 \\\\ & \Rightarrow n(p \cap q)=2! \end{array} $
$ \begin{aligned} & \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\\\ & = 5 !+4 !-2 !=120+24-2=142 \end{aligned} $
$\therefore n$ (neither string 143 nor string 2467)
$ =7 !-142=5040-142=4898 $
Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total number of persons, who participated in the tournament, is ___________.
Explanation:
According to the question, $2 \times{ }^n C_2 \times{ }^{n-2} C_2=840$
$ \begin{aligned} & \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\\\ & \Rightarrow \frac{n !}{2 !(n-2) !} \times \frac{(n-2) !}{(n-4) ! 2 !}=420 \\\\ & \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420 \end{aligned} $
Put $n=8$ satisfied the equation.
So, $n=8$
Hence, total number of players who participated
in the tournament $=2 \times 8=16$
The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is __________.
Explanation:
Here, vowels are E, I, U and consonants are N, R, S, V
$\therefore$ Required number of 4-letters words, with or without meaning,
each consisting of 2 vowels and 2 consonants
$ \begin{aligned} & ={ }^3 C_2 \times{ }^4 C_2 \times 4 ! \\\\ & ={ }^3 C_1 \times{ }^4 C_2 \times 24 \\\\ & =3 \times 6 \times 24=432 \end{aligned} $
The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is ___________.
Explanation:
Total ways without any restrictions :
There are $3^{20}$ ways to distribute the oranges to the 3 children.
Number of ways one child receives no orange :
Choose 1 child out of the 3 to not receive any orange in ${ }^3 C_1 = 3$ ways. Distribute 20 oranges to the remaining 2 children in $2^{20}$ ways. However, we've included the scenarios where the 2 children each get all the oranges. So, we subtract the 2 ways where one of the two remaining children gets all the oranges.
$ { }^3 C_1(2^{20} - 2) $
Number of ways two children receive no orange :
Choose 2 children out of the 3 to not receive any oranges in ${ }^3 C_2 = 3$ ways. The third child will receive all 20 oranges in $1^{20} = 1$ way.
$ { }^3 C_2 \times 1^{20} = 3 $
Number of ways
$=$ Total $-($ One child receive no orange + two child receive no orange)
$ \begin{aligned} & =3^{20}-\left({ }^3 C_1\left(2^{20}-2\right)+{ }^3 C_2 1^{20}\right) \\\\ & =3483638676 \end{aligned} $
Number of integral solutions to the equation $x+y+z=21$, where $x \ge 1,y\ge3,z\ge4$, is equal to ____________.
Explanation:
The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is ____________.
Explanation:
Units, place must be occupied by 4 and hence, at least one 4 must be there.
Possible combination of 4, 5, 9 are as follows :
| 4 | 5 | 9 | Total number of Number |
|---|---|---|---|
| 1 | 1 | 4 | ${{5!} \over {4!}} = 5$ |
| 1 | 4 | 1 | ${{5!} \over {4!}} = 5$ |
| 2 | 2 | 2 | ${{5!} \over {2!2!}} = 30$ |
| 3 | 0 | 3 | ${{5!} \over {2!3!}} = 10$ |
| 3 | 3 | 0 | ${{5!} \over {2!3!}} = 10$ |
| 4 | 1 | 1 | ${{5!} \over {3!}} = 20$ |
| 6 | 0 | 0 | ${{5!} \over {5!}} = 1$ |
Total = 5 + 5 + 30 + 10 + 10 + 20 + 1 = 81.
The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7, is ____________.
Explanation:
$A=$ Numbers divisible by 2
$B=$ Numbers divisible by 3
$C=$ Numbers divisible by 7
$n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=n(2)+n(3)-n(6)$
$n(A)=n(2)=100,102 \ldots ., 998=450$
$n(B)=n(3)=102,105, \ldots ., 999=30$
$n(A \cap B)=n(6)=102,108, \ldots ., 996=150$
$n(2$ or 3$)=450+300-150=600$
Now,
$n(\mathrm{~A} \cap C)=n(14)=112,126, \ldots ., 994=64$
$n(A \cap B \cap C)=n(42)=126,168, \ldots ., 966=21$
$n(B \cap C)=n(21)=105,126, \ldots \ldots, 987,=43$
$n(2$ or 3 not by 7$)=600-[64+43-21]$
$=514$
The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ___________.
Explanation:
Consonants : S,S,S,S,N,N,T
$ \therefore $ Total number of ways in which vowels come together
$=\frac{8 !}{4 ! 2 !} \times \frac{6 !}{3 ! 2 !}=50400$
The number of matrices A such that the sum of all entries is a prime number $\mathrm{p} \in(2,13)$ is __________.
Explanation:
Such that $\Sigma a_{i i}=3,5,7$ or 11
Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.
Then total number of possible matrices $ =4+12+4 $ $=20$
For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.
$\therefore $ Total possible matrices $=12+12+12+12+4=52$
For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.
$\therefore $ Total possible matrices $=80$
For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.
$\therefore $ Total number of matrices $=52$
$\therefore $ Total required matrices $=20+52+80+52$ $ =204 $
then $n^{2}+n+15$ is equal to :
Explanation:
$\frac{(2 n+1) 2 n}{(n+2)(n+1) n}=\frac{11}{21}$
$84 n+42=11\left(n^{2}+3 n+2\right)$
$11 n^{2}-51 n-20=0$
$(n-5)(11 n+4)=0$
$n=5, \frac{-4}{11}$ (Rejected $)$
$n^{2}+n+15=45$
Let 5 digit numbers be constructed using the digits $0,2,3,4,7,9$ with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is __________.
Explanation:
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to ____________.
Explanation:
$=\frac{2799-1002}{3}+1=600$
Numbers which are divisible by 11 (4 digit) and less than or equal to 2800
$=\frac{2794-1001}{11}+1=164$
Numbers which are divisible by 33 (4 digit) and less than or equal to 2800
$=\frac{2772-1023}{33}+1=54$
$\therefore$ Total numbers = $600+164-54=710$
seven digits 1, 2, 2, 2, 3, 3, 5 is ____________.
Explanation:
$.......1 \to {{6!} \over {2!3!}} = 60$
$.......3 \to {{6!} \over {3!}} = 120$
$.......5 \to {{6!} \over {3!2!}} = 60$
Total = 240
Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5, and are divisible by 15, is equal to ___________.
Explanation:
We have to make 4 digit numbers using the
digits, 1, 2, 3 and 5.
The unit digit of the 4 digit number will be 5.
Now, the sum (x + y + z) should be of the (3p + 1).
Therefore, the possible cases are
(x, y, z) = (1, 1, 5), (1, 1, 2), (2, 2, 3), (2, 3, 5), (3, 3, 1) and
(5, 5, 3).
So, total arrangements are
For $(1,1,5) \rightarrow \frac{3 !}{2 !}=3 ;$
For $(1,1,2) \rightarrow \frac{3 !}{2 !}=3 ;$
For $(2,2,3) \rightarrow \frac{3 !}{2 !}=3 ;$
For $(2,3,5) \rightarrow 3 !=6 ;$
For $(3,3,1) \rightarrow \frac{3 !}{2 !}=3 ;$
For $(5,5,3) \rightarrow \frac{3 !}{2 !}=3 ;$
So, total number of arrangements $=3+3+3+6+3+3=$ 21
The total number of 4-digit numbers whose greatest common divisor with 54 is 2, is __________.
Explanation:
$\gcd (a,54) = 2$ when a is a 4 digit no.
And $54 = 3 \times 3 \times 3 \times 2$
So, $a=$ all even no. of 4 digits $-$ Even multiple of 3 (4 digits)
$ = 4500 - 1500$
$ = 3000$
If all the six digit numbers $x_1\,x_2\,x_3\,x_4\,x_5\,x_6$ with $0< x_1 < x_2 < x_3 < x_4 < x_5 < x_6$ are arranged in the increasing order, then the sum of the digits in the $\mathrm{72^{th}}$ number is _____________.
Explanation:
23 $\ldots\ldots\ldots\ldots\ldots\rightarrow{ }^{6} C_{4}=\frac{15}{71}$
$72^{\text {th }}$ number $=245678$
Sum $=32$
Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is ____________.
Explanation:
Descending order means from highest to lowest.
$\therefore$ Descending order of digits 1, 2, 3, 5, 7 is = 7, 5, 3, 2, 1.
(1) Number starts with 7 :
$\therefore$ Total possible numbers start's with 7 are = 5C1 $\times$ 5C1 $\times$ 5C1 $\times$ 5C1 = 54
(2) Number of starts with 5 :
$\therefore$ Total possible numbers start's with 5 are = 5C1 $\times$ 5C1 $\times$ 5C1 $\times$ 5C1 = 54
(3) Number starts with 37 :
$\therefore$ Total possible numbers starts with 3 and 7 are = 5C1 $\times$ 5C1 $\times$ 5C1 = 53
(4) Number starts with 357 :
$\therefore$ Total possible numbers starts with 357 are = 5C1 $\times$ 5C1 = 52
(5) Number starts with 355 :
$\therefore$ Total possible numbers starts with 355 are = 5C1 $\times$ 5C1 = 52
(6) Number starts with 3537 :
$\therefore$ Total possible numbers starts with 3537 are = 5C1 = 5
(7) Number starts with 3535 :
$\therefore$ Total possible numbers starts with 3535 are = 5C1 = 5
(8) Next number is 35337 which is the required number.
$\therefore$ Position of the number 35337 is = 54 + 54 + 53 + 52 + 52 + 5 + 5 + 1 = 1436
A triangle is formed by X-axis, Y-axis and the line $3x+4y=60$. Then the number of points P(a, b) which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is ____________.
Explanation:
$(1,1)(1,2)-(1,14) \Rightarrow 14$ pts.
If $x=2, y=\frac{27}{2}=13.5$
$(2,2)(2,4) \ldots(2,12) \quad \Rightarrow 6$ pts.
If $\mathrm{x}=3, \mathrm{y}=\frac{51}{4}=12.75$
$(3,3)(3,6)-(3,12) \Rightarrow 4$ pts.
If $x=4, y=12$
$(4,4)(4,8) \quad \Rightarrow 2$ pts.
If $x=5 . y=\frac{45}{4}=11.25$
$(5,5),(5,10) \Rightarrow 2$ pts.
If $\mathrm{x}=6, \mathrm{y}=\frac{21}{2}=10.5$
$(6,6) \quad \Rightarrow 1 \mathrm{pt}$.
If $x=7, y=\frac{39}{4}=9.75$
$(7,7) \Rightarrow 1 \mathrm{pt}$.
If $x=8, y=9$
$(8,8) \quad \Rightarrow 1$ pt.
If $\mathrm{x}=9 \mathrm{y}=\frac{33}{4}=8.25 \Rightarrow$ no $\mathrm{pt}$.
Total $=31$ pts.
Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is ____________
Explanation:
Case I: 3 orange $+1$ red apple $+1$ white apple
$ ={ }^{8} C_{3} \times{ }^{7} C_{1} \times{ }^{5} C_{1}=1960 $
Case II : 2 oranges $+2$ red apples $+1$ white apple.
$ ={ }^{8} C_{2} \times{ }^{7} C_{2} \times{ }^{5} C_{1}=2940 $
Case III : 2 oranges $+1$ red apples $+2$ white apple.
$ \begin{aligned} & ={ }^{8} C_{2} \times{ }^{7} C_{1} \times{ }^{5} C_{2} \\\\ & =1960 \\\\ \text { Total } & =1960+2940+1960 \\\\ & =6860 \end{aligned} $
Let $x$ and $y$ be distinct integers where $1 \le x \le 25$ and $1 \le y \le 25$. Then, the number of ways of choosing $x$ and $y$, such that $x+y$ is divisible by 5, is ____________.
Explanation:
Possible cases are
$\begin{array}{llc}x & y & \text { Number of ways } \\ 5 \lambda(5,10,15,20,25) & 5 \lambda(5,10,15,20,25) & 20 \\ 5 \lambda+1(1,6,11,16,21) & 5 \lambda+4(4,9,14,19,24) & 25 \\ 5 \lambda+2(2,7,12,17,22) & 5 \lambda+3(3,8,13,18,23) & 25 \\ 5 \lambda+3(3,8,13,18,23) & 5 \lambda+2(2,7,12,17,22) & 25 \\ 5 \lambda+4(4,9,14,19,24) & 5 \lambda+1(1,6,11,16,21) & 25\end{array}$
Total number of ways $=20+25+25+25+25=120$
Note : In first case total number of ways = 20 as in the question given that the chosen $x$ and $y$ are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}.
Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25.
$ \therefore $ Total number of ways in the first case = 5 $ \times $ 4 = 20.
A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is __________
Explanation:
Among 12 courses, 5 courses are of language.
$\therefore$ Remaining 7 are different courses.
Now, number of ways to select 5 courses where at most 2 language courses present.
| Language | Different | Number of ways | |
|---|---|---|---|
| Case 1 | 0 | 5 | ${}^5{C_0} \times {}^7{C_5}$ |
| Case 2 | 1 | 4 | ${}^5{C_1} \times {}^7{C_4}$ |
| Case 3 | 2 | 3 | ${}^5{C_2} \times {}^7{C_3}$ |
$\therefore$ Total number of ways
$ = {}^5{C_0} \times {}^7{C_5} + {}^5{C_1} \times {}^7{C_4} + {}^5{C_2} \times {}^7{C_3}$
$ = 546$
The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ______________.
Explanation:
Here, even digits are 2 and 4.
Number of digit "2" presents = 2
Number of digit "4" presents = 2
$\therefore$ Total even digits = 4
$\therefore$ Total 4 even places presents.
Number of ways to put those 4 digits in those 4 places $ = {{{}^4{C_4} \times 4!} \over {2!2!}} = {{4!} \over {2!2!}}$
Now, remaining 5 digits (three 1 and two 3) can be put in those 5 places in $ = {{{}^5{C_5} \times 5!} \over {3!2!}} = {{5!} \over {3!2!}}$ ways.
$\therefore$ Total possible 9 digit numbers
$ = {{4!} \over {2!2!}} \times {{5!} \over {3!2!}} = 60$
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