5 Mark Each

Question 15 Marks

Using all the digits $2, 5, 7, 8$, four digit numbers are to be formed.
$(1)$ How many such numbers can be made ?
$(2)$ How many of them will be even numbers ?
$(3)$ How many of them will be divisible by $5 ?$
$(4)$ How many of them will be greater than $5000 ?$

Answer

Using all the digits $2, 5, 7, 8$ i.e. $4$ digits, total number 0f $4$ digit
numbers that can be formed is${ }^{4} \mathrm{P}_{4}$
$ \therefore \text { Total permutations } ={ }^{4} \mathrm{P}_{4}$
$ =4 !$
$ =24 $

(2)

If even numbers are to be formed using all the digit $2,5,7,8$ then digit $2$ or $8$ should be in unit place.
This arrangement can be done in ${ }^{2} P_{1}$ ways. After arranging $2$
or $8$ in unit place, remaining $3$ digits can be arranged in remaining $3$ places in ${ }^{3} P_{3}$ ways.
$ \therefore \text { Total Permutations } ={ }^{2} \mathrm{P}_{1} \times{ }^{3} \mathrm{P}_{3}$
$ =2 \times 3 !$
$ =2 \times 6$
$ =12 $
$(3)$

If numbers divisible by $5$ are to be formed using all the digits $2,5,7,8$ then digit $5$ should be in unit place.
This arrangement can be done in ${ }^{1} P_{1}$ ways. After arranging $5$ in unit place,
remaining $3$ digits can be arranged in remaining $3$ places in ${ }^{3} \mathrm{P}_{3}$ ways.
$ \therefore \text { Total Permutations } ={ }^{1} \mathrm{P}_{1} \times{ }^{3} \mathrm{P}_{3}$
$ =1 ! \times 3 !$
$ =1 \times 6$
$ =6 $
$(4)$

If numbers greater then $5000$ are to be formed using all digits $2,5,7,8$
then digit $5,7$ or $8$ should be in thousands place (first place).
This arrangement can be done in ${ }^{3} P_{1}$ ways.
After arranging $5,7$ or $8$ in thousands place,
remaining $3$ digits can be arranged in remaining $3$ places in ${ }^{3} \mathrm{P}_{3}$ ways :
$ \therefore \text { Total Permutations } ={ }^{3} P_{1} \times{ }^{3} P_{3}$
$ =3 \times 3 !$
$ =3 \times 6$
$ =18 $

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Question 25 Marks

There are $6$ engineers and $4$ managers in a company. In how many ways can a committee of $5$ members be made such that
$(1)$ there are at least $2$ managers ?
$(2)$ there are at the most $2$ engineers ?
$(3)$ engineers are in majority ?

Answer

$(l)$ in a committee of $5$ members, selection of at least $2$ managers can be done in following ways :

$ \therefore \text { Total Combinations } =\left({ }^{4} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{3}\right)+\left({ }^{4} \mathrm{C}_{3} \times{ }^{6} \mathrm{C}_{2}\right)+\left({ }^{4} \mathrm{C}_{4} \times{ }^{6} \mathrm{C}_{1}\right)$
$ =(6 \times 20)+(4 \times 15)+(1 \times 6)$
$ =120+60+6$
$ =186 $
$(2)$ In a committee of $5$ persons, selection of at the most $2$ engineers can be done in following ways:
$ \therefore \text { Total Combinations } =\left({ }^{6} \mathrm{C}_{2} \times{ }^{4} \mathrm{C}_{3}\right)+\left({ }^{6} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{4}\right)$
$ =(15 \times 4)+(6 \times 1)$
$ =60+6$
$ =66 $
$(3)$ The selection of committe of $5$ members such that engineers are in majority can be done in following ways :

$ \therefore \text { Total Combinations } =\left({ }^{6} \mathrm{C}_{5} \times{ }^{4} \mathrm{C}_{0}\right)+\left({ }^{6} \mathrm{C}_{4} \times{ }^{4} \mathrm{C}_{1}\right)+\left({ }^{6} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{2}\right)$
$ =(6 \times 1)+(15 \times 4)+(20 \times 6)$
$ =6+60+120$
$ =186 $

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Question 35 Marks

Arrangements are made using all the letters of the word $\text{YOUNG}$. If all these arrangements are arranged in the order of dictionary, what will be the rank of the word $\text{YOUNG ?}$

Answer

There are $5$ letters $\mathrm{Y}, \mathrm{O}, \mathrm{U}, \mathrm{N}, \mathrm{G}$ in the word $\mathrm{\text{YOUNG}}$ which can be arranged in ${ }^{5} \mathrm{P}_{5}=5 !=120$ ways.
Now, we have to obtain the order of the word $\text{YOUNG}$ from all 120 arrangements as per dictionary order.
Alphabetical order of all letters of the word $\text{YOUNG}$ will be $\text{G, N, O, U, Y}$
Arrangements with $\mathrm{G}$ in the first place will be ${ }^{1} \mathrm{P}_{1} \times{ }^{4} \mathrm{P}_{4}=24$.
Arrangements with $\mathrm{N}$ in the first place will be ${ }^{1} \mathrm{P}_{1} \times{ }^{4} \mathrm{P}_{4}=24$.
Arrangements with $\mathrm{O}$ in the first place will be ${ }^{1} \mathrm{P}_{1} \times{ }^{4} \mathrm{P}_{4}=24$.
Arrangements with $U$ in the first place will be ${ }^{1} P_{1} \times{ }^{4} P_{4}=24$.
Arrangements with $\mathrm{Y}$ in the first place and $\mathrm{G}$ in the second place will be ${ }^{1} \mathrm{P}_{1} \times{ }^{1} \mathrm{P}_{1} \times{ }^{3} \mathrm{P}_{3}=6$
Arrangements with $Y$ in the first place and $N$ in the second place will be ${ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{3} P_{3}=6$
Arrangements with $\mathrm{Y}$ in the first place, $\mathrm{O}$ in the second place and $\mathrm{G}$ in the third place will be ${ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{2} P_{2}=2$
Arrangements with $\mathrm{Y}$ in first place, $\mathrm{O}$ in second place and
$\mathrm{N}$ in the third place will be ${ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{2} P_{2}=2$.
Arrangements with $\mathrm{Y}$ in first place, $\mathrm{O}$ in the second place,
$\mathrm{U}$ in the third place and $\mathrm{G}$ in the fourth place will be ${ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{1} P_{1} \times{ }^{\prime} P_{1} \times{ }^{1} P_{1}=1$.
Thereafter, the word $\text{YOUNG}$ comes which itself takes $1$ position.
$\therefore$ Dictionary order of the word $\text{YOUNG}$ $=24+24+24+24+6+6+2+2+1+1=114$

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Question 45 Marks

Putting $x=2$ in the expansion of $(1+x)^{4}$ verify both side.

Answer

$\mathrm{LHS}=\mathrm{RHS}$

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Question 55 Marks

Find the Value of the following using binomial expansion: $(2+\sqrt{5})^{5}+(2-\sqrt{5})^{5}$

Answer

$1364$

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Question 65 Marks