2 Marks Questions

Question 12 Marks

From a class of 40 students, in how many ways can five students be chosen for an excursion party.

Answer

From a class of 40 students, five students can be chosen for an excursion party in ${ }^{40} C_5$ ways
$
\begin{array}{l}
=\frac{40!}{5!(40-5)!} \\
=\frac{40!}{5!35!}=\frac{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36.35!}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 35!} \\
=658,008
\end{array}
$

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Question 22 Marks

18 mice were placed in two experimental groups and one control group with all groups equally large. In how many ways can the mice be placed into three groups?

Answer

It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
$\therefore$ Required arrangements $=$
$
\frac{\text { Total arrangement }}{\text { Equally likely arrangement }}=\frac{18!}{6!6!6!} .
$

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Question 32 Marks

In how many ways can the letter of the word "PENCIL" be arranged so that I is always next to L .

Answer

There are 6 letters in the word "PENCIL". Consider LI as one letter. Now, 5 letters (P, E, N, C, ... LI) can be arranged in ${ }^5 P_5=5!=120$ ways.
Hence, the total number of ways in which I is always next to L is 120 .

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Question 42 Marks

How many ways are there to arrange the letters of the word "GARDEN" with the vowels in alphabetical order ?

Answer

Total number of ways in which all letters of the word GARDEN can be arranged $=6!=720 $
There are only two vowels in the word, A and E . When A at the first place, E can be occupy any of the remaining 5 places.
So, total arrangements $=5 \times 4$ !
When A in the second place, E can occupy any of 4 places,
So, total arrangements $=4 \times 4$ !
Repeat this process until A occupies the last but one place. A cannot occupy the last place.
Hence, total number of arrangement is
$(5+4+3+$ $2+1) \times 4$ !
$=15 \times 24=360$

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Question 52 Marks

In how many ways can a student choose a program of 5 courses, if 9 courses are available and 2 specific courses are compulsory for every student?

Answer

Total number of available courses $=9$
Out of these 5 courses have to be chosen. But it is given that 2 courses are compulsory for every student i.e., you have to choose only 3 courses instead of 5 , out of 7 instead of 9 .
It can be done in ${ }^7 C_3$ ways $=\frac{7 \times 6 \times 5}{6}=35$ ways.

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Question 62 Marks

In how many ways 7 pictures can be hanged on 9 pegs?

Answer

7 pictures have options of 9 pegs. So, they can be selected in ${ }^9 C_7$ ways.
Now, these pictures can arrange themselves in $7!$ ways.
$\therefore$ Number of ways by which 7 pictures can be hanged on 9 pegs $={ }^9 C_7 \times 7!$

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Question 72 Marks

Find the number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men.

Answer

Given,
Number of men $=4$
Number of women $=6$
It is given that committee includes at least two men and exactly twice as many women as men.
So, we can select either 2 men and 4 women or 3 men and 6 women.
$\therefore$ Required number of committee formed
$
\begin{array}{l}
={ }^4 C_2 \times{ }^6 C_4+{ }^4 C_3 \times{ }^6 C_6 \\
=6 \times 15+4 \times 1 \\
=94
\end{array}
$

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Question 82 Marks

Find the number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line.

Answer

Total number of triangles formed 12 points taking 3 at a time $={ }^{12} C_3$
But any three points selected from given seven collinear points does not from triangle.
Number of ways of selecting three points from seven collinear points $={ }^7 C_3$
$\begin{aligned} \therefore \text { Required number of triangles } & ={ }^{12} C_3-{ }^7 C_3 \\ & =220-35 \\ & =185 .\end{aligned}$

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Question 92 Marks

A gentleman has 6 friends to invite. In how many ways can be send invitation card to them if he has three servants to carry the cards ?

Answer

Given, that a gentleman has 6 friends to invite but he send invitations to them if he has 3 servants.
He send to invitations to one of his friends in 3 different ways,
Similarly, he send invitations to them if has 3 servants $=3 \times 3 \times 3 \times 3 \times 3 \times 3=3^6=729$.

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Question 102 Marks

How many different words can be formed by using all the letters of word 'SCHOOL'?

Answer

Since, 'SCHOOL' has 6 letters, out of these 6 letters there is $2 O ^{\prime} s$.
Hence number of permutations
$
\begin{array}{l}
=\frac{6!}{2!}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2} \\
=360
\end{array}
$

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Question 112 Marks

How many different words (with or without meaning) can be made using all the vowels at a time?

Answer

There are 5 vowels in 26 alphabets.
Hence, using all 5 vowels at a time, number of different words (with or without meaning) can be made are $=5!=5 \times 4 \times 3 \times 2 \times 1=120$.

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Question 122 Marks

How many numbers are there between 99 and 1000 having 7 in the unit place?

Answer

The number between 99 and 1000 are all three digit numbers. If we fix 7 at unit place, then middle digit can be any one of the 10 digits from 0 to 9 . The digit in hundred's place can be any of the 9 digits from 1 to 9 . Therefore, by the fundamental principle of counting, there are $10 \times 9=90$ numbers between 99 and 1000 having 7 in the units places.

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Applied Maths 2 Marks Questions

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