4 Marks Questions

Question 14 Marks

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if (i) they can be of any colour. (ii) two must be white and two red. (iii) they must all be of the same colour.

Answer

Total number of marbles $=6$ white +5 red $=1$ marbles
(i) If they can be of any colour means we have to select 4 marbles out of 11.
$\therefore$ Required Number of ways $={ }^{11} C_4=\frac{11!}{7!4!}=330$
(ii) If two must be white, then selection will be ${ }^6 C _2$ and two must be red, then selection will be ${ }^5 C_2$
$\therefore$ Required number of ways $={ }^6 C_2 \times{ }^5 C_2=15 \times 10$
$=150$
(iii) If they all must be same colour, then selection of 4 white marbles out of $6={ }^6 C_4$ and selection of 4 red marble out of $5={ }^5 C_4$
$\therefore$ Required number of ways $={ }^6 C_2+{ }^5 C_4=15+5$
$=20$

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Question 24 Marks

From 6 different novels and 3 different dictionaries, 4 novels and a dictionary is to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then find the number of such arrangements.

Answer

The number of ways in which 4 novels can be selected from 6 different novels $={ }^6 C_4=15$
The number of ways in which 1 dictionary can be selected out of 3 different dictionaries $={ }^3 C_1=3 \quad 1$
Now, it is given that dictionary is always in the middle.
Hence, the arrangement look like M M D M M
$\therefore$ Dictionary is always in the middle as the novels are different, hence they can be arranged in 4! ways.
Hence, total number of such arrangement
$
\begin{array}{l}
=15 \times 3 \times 4! \\
=15 \times 3 \times 4 \times 3 \times 2 \times 1 \\
=1080
\end{array}
$

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Question 34 Marks

How many different products can be obtained by multiplying two or more of the numbers $2,5,6,7$, $9 ?$

Answer

The given numbers are $2,5,6,7,9$.
The numbers of different products when 2 or more is taking = the number of ways of taking product of 2 numbers + number of ways of taking product of 3 numbers + numbers of ways of taking product of 4 numbers + number of ways of taking 5 together 2
$\begin{array}{l}={ }^5 C_2+{ }^5 C_3+{ }^5 C_4+{ }^5 C_5 \\ =\frac{5!}{3!2!}+\frac{5!}{2!3!}+\frac{5!}{1!4!}+\frac{5!}{0!5!}\end{array}$
$\begin{array}{l}=\frac{5 \cdot 4}{2 \cdot 1}+\frac{5 \cdot 4}{2 \cdot 1}+\frac{5}{1}+1 \\ =10+10+5+1 \\ =26 .\end{array}$

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Question 44 Marks

Find the number of all possible arrangements of the letters of the word "MATHEMATICS" taken four form at a time.

Answer

The word MATHEMATICS consists of 11 letters: (M, M), (A, A), (T, T), H, E, L, C, S
Case 1 : In this case 2 similar and 2 similar letters are selected, number of arrangements
$
\begin{array}{l}
={ }^3 C_2 \times \frac{4!}{2!2!} \\
=18
\end{array}
$

Case 2 : In this case 2 similar and 2 different letters are selected, number of arrangements
$
={ }^3 C_1 \times{ }^7 C_2 \times \frac{4!}{2!}=756
$

Case 3 : In this case all 4 letters selected are different, number of arrangements $={ }^8 C_4 \times 4!=1680$.
Therefore, total number of arrangements
$=18+756+1680=2454 \cdot $

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Question 54 Marks

A committee of 7 has to be formed out of 9 boys and 4 girls. In how many ways can be this be done when the committee consists of
(i) exactly 3 girls
(ii) At most 3 girls ?

Answer

A committee of 7 has to be formed from $9 B$ and $4 G$.
(i) exactly 3 girls
$={ }^9 C_4 \times{ }^4 C_3$
$\begin{array}{l}=\frac{\lfloor{9}}{\lfloor 4 \lfloor 5} \times \frac{\lfloor 4}{\lfloor 3 \lfloor 1}=\frac{9 \times 8 \times 7 \times 6 \times \lfloor{5}}{3 \times 2 \times \lfloor{5}} \\ =72 \times 7=504 .\end{array}$

(ii) at most 3 girls
(a) No girls and 7 boys
(b) 1 girl and 6 boys
(c) 2 girls 5 boys
(d) 3 girls and 4 boys
$\therefore$ The committee consists of at most 3 girls
$
\begin{array}{l}
={ }^4 C_0 \times{ }^9 C_7+{ }^4 C_1 \times{ }^9 C_6+{ }^4 C_2 \times{ }^9 C_5+{ }^4 C_3 \times{ }^9 C_4 \\
=36+336+1296+504 \\
=2172
\end{array}
$

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Question 64 Marks

Find the number of word with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in dictionary, what will be the $50^{\text {th }}$ word?

Answer

Number of words made from AGAIN $=\frac{5!}{2!}=60$
To get the number of words starting with $A$, we fix the letter $A$ at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutation of 4 different things taken 4 at a time.
Hence, the number of words starting with $A =4!=$ 24
Then, starting with G , the number of words $=\frac{4!}{2!}$ $=12$.
As after placing $G$ at the extreme left position, we are left with the letters, $A, A, I$ and $N$. Similarly, there are 12 words starting with the next letter $I$.
Total number of words so far obtained $=24+12+$ $12=48$
The $49^{\text {th }}$ word is NAAGI. The $50^{\text {th }}$ word is NAAIG.

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