5 Marks Questions

Question 15 Marks

If ${ }^n C_{r-1}=36,{ }^n C_r=84$ and ${ }^n C_{r+1}=126$, then find the value of ${ }^r C_2$.

Answer

$\begin{array}{rlrl}\text { Given, } { }^n C_{r-1} =36\quad \ldots(i) \\ \Rightarrow\quad\quad { }^n C_r =84\quad \ldots(ii) \\ \Rightarrow\quad { }^n C_{r+1} =126\quad \ldots(iii)\end{array}$
On dividing Eq. (i) by (ii), we get
$
\frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{36}{84} \quad\left[\begin{array}{l}
\because{ }^n C_r=\frac{n!}{(n-r)!r!} \\
\text { and } n!=n(n-1)!
\end{array}\right]
$
$
\begin{array}{l}
\Rightarrow \quad \frac{n!}{(r-1)!\{n-(r-1)\}!} \cdot \frac{r!(n-r)!}{n!}=\frac{3}{7} \\
\Rightarrow \frac{1}{(r-1)!(n-r+1)!} \cdot \frac{r(r-1)!(n-r)!}{1}=\frac{3}{7} \\
\Rightarrow \quad \frac{1 \cdot r}{(n-r+1)} \cdot(n-r)!=\frac{3}{7} \\
\Rightarrow \quad \frac{r}{n-r+1}=\frac{3}{7} \\
\Rightarrow \quad 7 r=3 n-3 r+3 \\
\Rightarrow \quad 10 r-3 n=3\quad \ldots(iv)
\end{array}
$
On dividing Eq. (ii) by Eq. (iii), we get
$
\frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{84}{126}
$
$
\begin{array}{l}
\Rightarrow \quad \frac{n!}{r!(n-r)!} \cdot \frac{(r+1)!(n-r-1)!}{n!}=\frac{14}{21} \\
\Rightarrow \frac{1!}{r!(n-r)(n-r-1)!} \cdot \frac{(r+1)!(n-r-1)!}{n!}=\frac{2}{3} \\
\Rightarrow \quad \frac{r+1}{n-r}=\frac{2}{3} \\
\Rightarrow \quad 3 r+3=2 n-2 r \\
\Rightarrow \quad 2 n-5 r=3\quad \ldots(v)
\end{array}
$
On multiplying Eq. (iv) by 2 and Eq. (v) by 3, we get
$
\begin{array}{l}
20 r-6 n=6\quad \ldots(vi) \\
6 n-15 r=9\quad \ldots(vii)
\end{array}
$
On adding Eq. (vi) and (vii),
$
5 r=15 r=3
$
From Eq. (v) $2 n=3+15$
$
\begin{aligned}
\Rightarrow 2 n =18 \Rightarrow n=9 \\
\therefore { }^r C_2 ={ }^3 C_2=\frac{3!}{2!1!} \\
=\frac{3 \times 2!}{2!}=3
\end{aligned}
$

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Question 25 Marks

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if a team has
(i) no girl
(ii) at least 3 girls
(iii) At least one boy and one girl

Answer

(i) no girl

	Total No.	No. to chosen	No. of ways to choose
Girls	4	0	${ }^4 C_0$
boys	7	5	${ }^7 C_5$

$\begin{aligned} \text { Total numbers of ways } & ={ }^4 C_0 \times{ }^7 C_5 \\ & =\frac{4!}{0!(4-0)!} \times \frac{7!}{5!(7-5)!} \\ & =\frac{4!}{4!} \times \frac{7!}{5!2!} \\ & =1 \times \frac{7 \times 6}{2} \\ & =21 .\end{aligned}$

(ii) at least 3 girls ?
Since, the team has to consist of at least 3 girls, the team can consist of
(a) 3 girls and 2 boys
(b) 4 girls and 1 boy

(a) 3 girls and 2 boys

	Total No.	No. to chosen	No. of ways to choose
Girls	4	3	${ }^4 C_3$
boys	7	2	${ }^7 C_2$

$\begin{aligned} \text { Numbers of ways selecting } & ={ }^7 C_2 \times{ }^4 C_3 \\ & =\frac{7!}{2!5!} \times \frac{4!}{3!1!} \\ & =\frac{7 \times 6}{2 \times 1} \times 4 \\ & =84 .\end{aligned}$

(b) 4 girls and 1 boys

	Total No.	No. to chosen	No. of ways to choose
Girls	4	4	${ }^4 C_4$
boys	7	1	${ }^7 C_1$

$
\begin{aligned}
\text { Numbers of ways selecting } & ={ }^7 C_1 \times{ }^4 C_4 \\
& =\frac{7!}{1!6!} \times \frac{4!}{4!0!} \\
& =7 \times 1 \\
& =7
\end{aligned}
$
$\therefore$ Total number of ways $=84+7=91$.

(iii) at least one girl and one boy?
A group giving at least one boy and one girl will consists of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girls
Number of ways of selecting 1 boy and 4 girls
$
={ }^7 C_1 \times{ }^4 C_4=7
$
Number of ways of selecting 2 boys and 3 girls
$
={ }^7 C_2 \times{ }^4 C_3=84
$
Number of ways of selecting 3 boys and 2 girls
$
={ }^7 C_3 \times{ }^4 C_2=210
$
Number of ways of selecting 4 boys and 1 girl
$
={ }^7 C_4 \times{ }^4 C_1=140
$
Hence, total number of ways
$
\begin{array}{l}
=7+84+210+140 \\
=441 \text { ways. }
\end{array}
$

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Question 35 Marks

How many words, with or without meaning can be made from the letters of the word 'MONDAY', assuming that no letter is repeated, if
(i) 4 letters are used at a time.
(ii) all letters are used at a time.
(iii) all letters are used but first letter is a vowel.

Answer

In the word MONDAY, all letters are different.
(i) Out of 6 different letters 4 letters can be selected in ${ }^6 P_4$ ways.
$\therefore$ Required number of words $={ }^6 P_4$
$
=\frac{6!}{(6-4)!}=\frac{6!}{2!}=\frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}=360 .
$

(ii) The word 'MONDAY' has 6 different letters.Number of ways taking 6 letters at a time $={ }^6 P_6$
$\therefore$ Required number of words $={ }^6 P_6$
$
=\frac{6!}{(6-6)!}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{0!}=720
$

(iii) First, we will fix the vowel.
In the word MONDAY, there are two vowels $O$ and $A$.
$\therefore$ First letter can be chosen by 2 ways.
Number of ways taking 5 different letters from remaining 5 letters $={ }^5 P_5$
$\therefore$ Required number of words$
={ }^5 P_5=\frac{5!}{(5-5)!}=\frac{5!}{0!}=5 \times 4 \times 3 \times 2 \times 1=120
$
Hence, total number of ways $=2 \times 120=240 .$

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Question 45 Marks

In how many ways can be letters of the word PERMUTATIONS be arranged if the :
(i) Words start with $P$ and end with $S$,
(ii) Vowels are all together,
(iii) There are always 4 letters between $P$ and $S$ ?

Answer

(i) Letters between $P$ and $S$ are ERMUTATION.
These 10 letters having $T$ two times. These letters can be arranged in $\frac{10!}{2!}=1814400$ ways.

(ii) There are 12 letters in the word PERMUTATIONS which have $T$ two times.
Now, the vowels A, E, I, O, U are taken together.
Let it be considered in one block.
The letters of vowels can be arranged in 5! ways.
Thus, there are 7 letters and 1 block of vowels with T two times.$
\begin{array}{l}
\therefore \text { Required number of arrangements }=\frac{8!}{2!} \times 5! \\
=24,19,200
\end{array}
$

(iii) There are 12 letters to be arranged in 12 places
$
\begin{array}{|l|l|l|l|l|l|l|l|l||l|l|}
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
\hline
\end{array}
$
These 12 letters are to filled in 12 places shown above. $P$ may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently $S$ may be filled up at place No.$6,7,8,9,10,11,12$ leaving four places in between. Now $P$ and $S$ may be filled up in 7 ways.
Similarly, $S$ and $P$ may be filled in 7 ways.
Remaining ten letters having two $T^{\prime}$ s
$\therefore$ These can be arrange $\frac{10!}{2!}$
Hence, word PERMUTATIONS arrange when 4 letters between $P$ and $S$
$
\begin{array}{l}
=2 \times 7 \times \frac{10!}{2!} \\
=14 \times 1814400 \\
=25401600
\end{array}
$

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Question 55 Marks

In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back ? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats ?

Answer

We have 14 seats on two vans and there are 9 boys and 3 girls
i.e., total 12 peoples
The number of ways of arranging 12 peoples on 14 seats without restriction is ${ }^{14} P_{12}$.
Now,
$\quad \begin{aligned}{ }^{14} P_{12} & =\frac{14!}{2!}=\frac{14.13!}{2!} \\ & =7 \times 13!\text { ways }\end{aligned}$
Three girls can be seated together in back row on adjacent seats in the following ways
$1,2,3$ or $2,3,4$ of first van
$1,2,3$ or $2,3,4$ of second van
In each way the three girls can interchange among themselves in 3! ways
$\therefore$ Total number of ways in which 3 girls sit together in a back row $=4 \times 3!=24$ ways
$
\begin{array}{l}
9 \text { boys are to seated on } 11 \text { seats }={ }^{11} P_9=\frac{11!}{2!} \\
\text { Hence, the total number of ways }=\frac{24 \times 11!}{2!} \\
=12 \times 11!
\end{array}
$

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Question 65 Marks

If all the word "MOTHER" are written in all possible orders and the word so formed are arranged in a dictionary order, then find the rank of word 'MOTHER' ?

Answer

The given word is a MOTHER. In dictionary, the words are arranged in alphabetical order. The words beginning with $E , H , M , O , R$ and T are written in order.
Number of words starting with $E =5!=120$
Number of words starting with $H =5!=120$
Number of words starting with $M=5!=120$
But one of these word is MOTHER.
Now, number of words with $ME =4!=24$
Number of words with $MH =4!=24$
Number of words with $MO =4!=24$,
But one of these words is MOTHER.
Now, number of words with $MOE =3!=6$
Number of words with $MOH =3!=6$
Number of words with $M O T =3!=6$,
But one of these words is MOTHER.
The first words beginning with MOT is MOTEHR.
The second words is MOTERH. The third words is MOTHER.
Therefore, rank of MOTHER $=2(120)+2(24)+$ $3(6)+3$
$\begin{array}{l}=240+48+18+3 \\ =309\end{array}$

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Question 75 Marks

If ${ }^n C_r:{ }^n C_{r+1}=1: 2$ and ${ }^n C_{r+1}:{ }^n C_{r+2}=2: 3$, determine the values of $n$ and $r$.

Answer

Given, $\frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{1}{2}$
$\begin{array}{l}\Rightarrow \quad \frac{\left(\frac{n!}{r!(n-r)!}\right)}{\left(\frac{n!}{(r+1)!(n-r-1)!}\right)}=\frac{1}{2} \\ \Rightarrow \quad \frac{(r+1)!(n-r-1)!}{r!(n-r)!}=\frac{1}{2}\end{array}$
$\begin{aligned} \Rightarrow & \frac{r+1}{n-r} =\frac{1}{2} \\ \Rightarrow & 2 r+2 =n-r \\ \Rightarrow & n =3 r+2\quad \ldots(i) \end{aligned}$
Similarly, $\frac{{ }^n C_{r+1}}{{ }^n C_{r+2}}=\frac{2}{3}$
$\begin{array}{l}\Rightarrow \frac{\left(\frac{n!}{(r+1)!(n-r-1)!}\right)}{\left(\frac{n!}{(r+2)!(n-r-2)!}\right)}=\frac{2}{3} \\ \Rightarrow \quad \frac{(r+2)!(n-r-2)!}{(r+1)!(n-r-1)!}=\frac{2}{3}\end{array}$
$\Rightarrow \quad \frac{r+2}{n-r-1}=\frac{2}{3}$
$\Rightarrow \quad 3 r+6=2 n-2 r-2$
$\Rightarrow \quad 2 n=5 r+8\quad \ldots(ii) $
Solving (i) and (ii), we get
$6 r+4=5 r+8$
$\begin{array}{ll}\Rightarrow & r=4 \\ \text { and } & n=14 .\end{array}$

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Applied Maths 5 Marks Questions

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