Solve the Following Question.(3 Marks)

Question 13 Marks

How many numbers formed using the digits $3,2,0,4,3,2,3$ exceed one million?

Answer

A number that exceeds one million is to be formed from the digits $3,2,0,4$, $3,2,3$.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also among the given numbers 2 repeats twice and 3 repeats thrice.
$\therefore$ Required number of numbers $=$ Total number of arrangements possible among these digits - number of arrangements of 7 digits which begin with 0.
$
\begin{aligned}
& =\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !} \\
& =\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !} \\
& =7 \times 6 \times 5 \times 2-6 \times 5 \times 2 \\
& =6 \times 5 \times 2(7-1) \\
& =60 \times 6 \\
& =360
\end{aligned}
$
$\therefore 360$ numbers that exceed one million can be formed with the digits $3,2,0$, $4,3,2,3$.

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Question 23 Marks

Find the value of $r$ if ${ }^{56} \mathrm{C}_{r+6}:{ }^{54} \mathrm{P}_{\mathrm{r}-1}=30800: 1$

Answer

$
\begin{array}{ll}
& { }^{56} \mathrm{P}_{\mathrm{r}+6}:{ }^{54} \mathrm{P}_{\mathrm{r}+3}=30800: 1 \\
\therefore \quad & \frac{{ }^{56} \mathrm{P}_{\mathrm{r}+6}}{{ }^{54} \mathrm{P}_{\mathrm{r}+3}}=\frac{30800}{1} \\
& \frac{56 !}{(56-\mathrm{r}-6) !}=30800 \\
\therefore \quad & \frac{54 !}{(54-\mathrm{r}-3) !} \\
\therefore \quad & \frac{56 !}{(56-\mathrm{r}-6) !} \times \frac{(54-\mathrm{r}-3) !}{54 !}=30800 \\
\therefore \quad & \frac{56 !}{(50-\mathrm{r}) !} \times \frac{(51-\mathrm{r}) !}{54 !}=30800 \\
\therefore \quad & \frac{56 \times 55 \times 54 !}{(50-\mathrm{r}) !} \times \frac{(51-\mathrm{r})(50-\mathrm{r}) !}{54 !}=30800 \\
\therefore \quad & 51-\mathrm{r}=\frac{30800}{56 \times 55} \\
\therefore \quad & 51-\mathrm{r}=10 \\
\therefore \quad \mathrm{r}=51-10 \\
\therefore \quad & \mathrm{r}=41
\end{array}
$

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Question 33 Marks

Find the differences between the largest values in the following: ${ }^{15} C_r-{ }^{11} C_r$

Answer

Greatest value of ${ }^{15} \mathrm{C}_{\mathrm{r}}$
Here $\mathrm{n}=15$, which is odd
Greatest value of ${ }^n C_r$ occurs at $r=\frac{n-1}{2}$ if $n$ is odd
$
\begin{array}{ll}
\therefore & \mathrm{r}=\frac{\mathrm{n}-1}{2} \\
\therefore & \mathrm{r}=\frac{15-1}{2}=7
\end{array}
$
$\therefore \quad$ Greatest value of ${ }^{15} \mathrm{C}_{\mathrm{r}}={ }^{15} \mathrm{C}_7=\frac{15 !}{7 ! 8 !}$
$
=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 !}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 8 !}
$
$=6435$
Also, for greatest value of ${ }^{11} \mathrm{C}_{\mathrm{r}}$ $\mathrm{n}=11$, which is odd
$
\begin{aligned}
\therefore \quad \mathrm{r}=\frac{11-1}{2} & =5 \\
\therefore \quad{ }^{11} \mathrm{C}_5 & =\frac{11 !}{5 ! 6 !} \\
& =\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{5 \times 4 \times 3 \times 2 \times 1 \times 6 !} \\
& =462
\end{aligned}
$
Difference between the greatest values of ${ }^{15} C_r$ and ${ }^{11} C_r=6435-462=5973$Find the differences between the largest values in the following: ${ }^{15} C_r-{ }^{11} C_r$

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Question 43 Marks

Find the differences between the largest values in the following: ${ }^{13} \mathrm{C}_{\mathrm{r}}-{ }^8 \mathrm{C}_{\mathrm{r}}$

Answer

Greatest value of ${ }^{13} \mathrm{C}_{\mathrm{r}}$
Here $\mathrm{n}=13$, which is odd
Greatest value of ${ }^n C_r$ occurs at $r=\frac{n-1}{2}$ if $n$ is odd
$
\begin{array}{ll}
\therefore & r=\frac{n-1}{2} \\
\therefore & r=\frac{13-1}{2}=6
\end{array}
$
$
\begin{aligned}
& \therefore \quad \text { Greatest value of }{ }^{13} \mathrm{C}_{\mathrm{r}}={ }^{13} \mathrm{C}_6=\frac{13 !}{6 ! 7 !} \\
&=\frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 !}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 7 !} \\
&=1716
\end{aligned}
$
Also, for greatest value of ${ }^8 \mathrm{C}_{\mathrm{r}}$ $\mathrm{n}=8$, which is even
$
\begin{aligned}
\therefore \quad r= & \frac{8}{2}=4 \\
\therefore \quad{ }^8 C_4 & =\frac{8 !}{4 ! 4 !} \\
& =\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !} \\
& =70
\end{aligned}
$
$\therefore$ Difference between the greatest values of ${ }^{13} \mathrm{C}_{\mathrm{r}}$ and ${ }^8 \mathrm{C}_{\mathrm{r}}=1716-70=1646$

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Question 53 Marks

Find the differences between the largest values in the following: ${ }^{14} \mathrm{C}_{\mathrm{r}}-{ }^{12} \mathrm{C}_{\mathrm{r}}$

Answer

Greatest value of ${ }^{14} \mathrm{C}_{\mathrm{r}}$
Here $n=14$, which is even
Greatest value of ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ occurs at $\mathrm{r}=\frac{n}{2}$ if $\mathrm{n}$ is even
$
\begin{array}{ll}
\therefore & r=\frac{n}{2} \\
\therefore & r=\frac{14}{2}=7
\end{array}
$
$
\begin{aligned}
& \therefore \quad \text { Greatest value of }{ }^{14} \mathrm{C}_{\mathrm{r}}={ }^{14} \mathrm{C}_7=\frac{14 !}{7 ! 7 !} \\
&=\frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 !}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 7 !} \\
&=3432
\end{aligned}
$
Also, for greatest value of ${ }^{12} \mathrm{C}_{\mathrm{r}}$ $\mathrm{n}=12$, which is even
$
\begin{aligned}
\therefore \quad r=\frac{12}{2} & =6 \\
\therefore \quad{ }^{12} \mathrm{C}_6 & =\frac{12 !}{6 ! 6 !} \\
& =\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6 !} \\
& =924
\end{aligned}
$
$\therefore$ Difference between the greatest values of ${ }^{14} C_r$ and ${ }^{12} C_r=3432-924=$ 2508
\mathrm{C}_{\mathrm{r}}$ $\mathrm{n}=12$, which is even
$
\begin{aligned}
\therefore \quad \mathrm{r}=\frac{12}{2} & =6 \\
\therefore \quad{ }^{12} \mathrm{C}_6 & =\frac{12 !}{6 ! 6 !} \\
& =\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6 !} \\
& =924
\end{aligned}
$
$\therefore$ Difference between the greatest values of ${ }^{14} C_r$ and ${ }^{12} C_r=3432-924=$ 2508

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Question 63 Marks

A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in the majority?

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Question 73 Marks

Find n and r if.: ${ }^n C_{r-1}:{ }^n C_r:{ }^n C_{r+1}=20: 35: 42$

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Question 83 Marks

Find n and r if.: ${ }^n P_r=720$ and ${ }^n C_{n-r}=120$

Answer

$
\begin{aligned}
& { }^n \mathrm{P}_{\mathrm{r}}=720 \\
& \therefore \quad \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=720 \\
& \text { Also, }{ }^n \mathrm{C}_{\mathrm{n}-\mathrm{r}}=120 \\
& \therefore \quad \frac{n !}{(n-r) !(n-n+r) !}=120 \\
& \therefore \quad \frac{n !}{r !(n-r) !}=120 \\
& \frac{\frac{n !}{(n-r) !}}{\frac{n !}{r !(n-r) !}}=\frac{720}{120} \\
& \therefore \quad r !=6 \\
& \therefore \quad \mathrm{r}=3 \\
&
\end{aligned}
$
Dividing (i) by (ii), we get
Substituting $r=3$ in (i), we get
$
\begin{aligned}
& \frac{n !}{(n-3) !}=720 \\
\therefore \quad & \frac{n(n-1)(n-2)(n-3) !}{(n-3) !}=720 \\
\therefore \quad & n(n-1)(n-2)=10 \times 9 \times 8 \\
\therefore \quad & n=10
\end{aligned}
$

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Question 93 Marks

Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N's together?

Answer

There are 6 letters in the word INDIAN in which I and N repeat twice.
Number of different words that can be formed using the letters of the word INDIAN =
$
\begin{aligned}
& \frac{6 !}{2 ! 2 !} \\
& =\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !} \\
& =180^2
\end{aligned}
$
$\therefore 180$ different words can be formed with the letters of the word INDIAN.
When two N's are together.
Let us consider the two N's as one unit.
They can be arranged with 4 other letters in $\frac{5 !}{2 !}$
$
\begin{aligned}
& =\frac{5 \times 4 \times 3 \times 2 !}{2 !} \\
& =60 \text { ways. }
\end{aligned}
$
$\therefore 2 \mathrm{~N}$ can be arranged in 1 way
$\therefore$ Total number of arrangements $=60 \times 1=60$ ways
$\therefore 60$ words are such that two N's are together.

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Question 103 Marks

Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements have two R's and two A's not together?

Answer

(i) There are 7 letters in the word ARRANGE in which A is repeated 2 times and $R$ is repeated 2 times
$\therefore$ The number of arrangements $=\frac{7 !}{2 ! 2 !}=1260$
(ii) A: set of words having $2 \mathrm{~A}$ together
$B$ : set of words having $2 R$ together
Number of words having both $A$ and both $R$ not together
$
\begin{aligned}
& =1260-n(A \cup B) \\
& =1260-[n(a)+n(B)-n(A \cap B)]
\end{aligned}
$
$\mathrm{n}(\mathrm{A})=$ number of ways in which (AA) $R, R, N, G, E$ are to be arranged
$
\therefore \mathrm{n}(\mathrm{A})=\frac{6 !}{2 !}=360
$
$\mathrm{n}(\mathrm{B})=$ number of ways in which (RR), A, A, N, G, E are to be arranged
$
\therefore \mathrm{n}(\mathrm{B})=\frac{6 !}{2 !}=360
$
$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=$ number of ways in which (AA), (RR), N, G, E are to be arranged
$
\therefore \mathrm{n}(\mathrm{A} \cap \mathrm{B})=5 !=120
$
Substituting $n(A), n(B), n(A \cap B)$ in (i), we get
Number of words having both $A$ and both $R$ not together
$
\begin{aligned}
& =1260-[360+360-120] \\
& =1260-600 \\
& =660
\end{aligned}
$

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Question 113 Marks

Find $m$ and $n$ if ${ }^{(m+n)} P_2=56$ and ${ }^{(m-n)} P_2=12$

Answer

$
\begin{aligned}
& \mathrm{m}^n \mathrm{P}_2=56 \\
& \therefore \frac{(\mathrm{m}+\mathrm{n}) !}{(\mathrm{m}+\mathrm{n}-2) !}=56 \\
& \therefore \frac{(m+n)(m+n-1)(m+n-2) !}{(m+n-2) !}=56 \\
& \therefore(\mathrm{m}+\mathrm{n})(\mathrm{m}+\mathrm{n}-1)=8 \times 7
\end{aligned}
$
Comparing on both sides, we get
$
\mathrm{m}+\mathrm{n}=8 \text {.....(i) }
$
Also ${ }^{\mathrm{m}-\mathrm{n}} \mathrm{P}_2=12$
$
\begin{aligned}
& \therefore \frac{(m-n) !}{(m-n-2) !}=12 \\
& \therefore \frac{(m-n)(m-n-1)(m-n-2) !}{(m-n-2) !}=12 \\
& \therefore(m-n)(m-n-1)=4 \times 3
\end{aligned}
$
Comparing on both sides, we get
$
m-n=4
$
Adding (i) and (ii), we get
$
\begin{aligned}
& 2 \mathrm{~m}=12 \\
& \therefore \mathrm{m}=6
\end{aligned}
$
Substituting $m=6$ in (ii), we get
$
\begin{aligned}
& 6-n=4 \\
& \therefore n=2
\end{aligned}
$

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Question 123 Marks

Find $n$ if $\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}=24: 1$

Answer

$
\begin{aligned}
& \frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}=24: 1 \\
& \therefore \quad \frac{(2 n) !}{7 !(2 n-7) !} \times \frac{4 !(n-4) !}{n !}=24 \\
& \therefore \quad \frac{(2 n)(2 n-1)(2 n-2)(2 n-3)(2 n-4)(2 n-5)(2 n-6)(2 n-7) !}{7 \times 6 \times 5 \times 4 !(2 n-7) !} \times \\
& \frac{4 !(n-4) !}{n(n-1)(n-2)(n-3)(n-4) !}=24 \\
& \therefore \quad \frac{(2 n)(2 n-1)(2 n-2)(2 n-3)(2 n-4)(2 n-5)(2 n-6)}{7 \times 6 \times 5} \\
& \text { MaharashtraBoardSolutions.in } \\
& \times \frac{1}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=24 \\
& \therefore \quad \frac{(2 n)(2 n-1) 2(n-1)(2 n-3) 2(n-2)(2 n-5) 2(n-3)}{7 \times 6 \times 5} \\
& \times \frac{1}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=24 \\
& \therefore \quad \frac{16(2 n-1)(2 n-3)(2 n-5)}{7 \times 6 \times 5}=24 \\
& \therefore \quad(2 n-1)(2 n-3)(2 n-5)=\frac{24 \times 7 \times 6 \times 5}{16} \\
& \therefore(2 n-1)(2 n-3)(2 n-5)=9 \times 7 \times 5 \\
&
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& 2 n-1=9 \\
& \therefore n=5
\end{aligned}
$

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Question 133 Marks

How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?

Answer

Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8
Number of such numbers = 3

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8
Ten’s place digit is selected from 2, 5, 7, 8.
∴ Ten’s place digit can be selected in 4 ways.
Unit’s place digit is anyone from 0, 1, 2, 5, 7, 8
∴ The unit’s place digit can be selected in 6 ways.
Using the multiplication principle,
the number of such numbers (repetition allowed) = 4 × 6 = 24

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8
100’s place digit is anyone from 1, 2, 5, 7, 8.
∴ 100’s place digit can be selected in 5 ways.
As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8
∴ 10’s place and unit’s place digits can be selected in 6 ways each.
Using multiplication principle,
the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180
All cases are mutually exclusive and exhaustive.
∴ Required number = 3 + 24 + 180 = 207

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Maths (commerce) Solve the Following Question.(3 Marks)

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