MCQ

MCQ 11 Mark

The number of ways in which a team of 11 players can be selected from 22 players always including 2 of them and excluding 4 of them is

A
${ }^{16} C_{11}$
B
${ }^{16} C_5$
✓
${ }^{16} C_9$
D
${ }^{20} C_9$

Answer

Correct option: C.

${ }^{16} C_9$

(C) ${ }^{16} C_9$
Explanation : Total number of players $=22$.
Since, given 2 players are always included and 4 players are always excluded or never included.
Therefore, total number of players $=22-2-4=16$ Now, we have to choose 11 players out of which 2 are included, then we have to choose only $11-2=$ 9 players.
$\therefore$ Required number of selections $={ }^{16} C _9$.

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MCQ 21 Mark

If ${ }^n C_{12}={ }^n C_8$, then $n$ is equal to

✓
20
B
12
C
6
D
30

Answer

Correct option: A.

(A) 20
Explanation : Given, ${ }^n C_{12}={ }^n C_8$
$
\Rightarrow \quad{ }^n C_{12}={ }^n C_{n-8} \quad\left[\because{ }^n C_r={ }^n C_{n-r}\right]
$
$\Rightarrow \quad 12=n-8$
$\Rightarrow \quad n=20$

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MCQ 31 Mark

The straight lines $l_1, l_2, l_3$ are parallel and lie in the same plane. A total number of $m$ points are taken on $l_1 ; n$ points on $l_2 ; k$ points on $l_3$, The maximum number of triangles formed with vertices at these points are

A
${ }^{(m+n+k)} C_3$
✓
${ }^{(m+n+k)} C_3-{ }^m C_3-{ }^n C_3-{ }^k C_3$
C
${ }^m C_3+{ }^n C_3+{ }^k C_3$
D
${ }^m C_3 \times{ }^n C_3 \times{ }^k C_3$

Answer

Correct option: B.

${ }^{(m+n+k)} C_3-{ }^m C_3-{ }^n C_3-{ }^k C_3$

(B) ${ }^{(m+n+k)} C_3-{ }^m C_3-{ }^n C_3-{ }^k C_3$
Explanation : Here, total number of points are $(m+n+k)$ which must give ${ }^{(m+n+k)} C_3$ number of triangles but $m$ points on line $l_1$ taking 3 points at a time gives ${ }^m C_3$ combinations which produce no triangle. Similarly, ${ }^n C_3$ and ${ }^k C_3$ number of triangles cannot be formed.
Therefore, the required number of triangles is
$
{ }^{(m+n+k)} C_3-{ }^m C_3-{ }^n C_3-{ }^k C_3
$

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MCQ 41 Mark

In how many ways a committee consisting of 3 men and 2 women, can be chosen from 7 men and 5 women?

A
45
✓
350
C
4200
D
230

Answer

Correct option: B.

350

(B) 350
Explanation: Out of the 7 men, 3 men can be chosen in ${ }^7 C_3$ ways and out of the 5 women, 2 women can be chosen in ${ }^5 C _2$ ways. Hence, the committee can be chosen in ${ }^7 C_3 \times{ }^5 C_2=350$ ways.

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MCQ 51 Mark

The sum of the digits in unit place of all the numbers formed with the help of $3,4,5$ and 6 taken all at a time is

A
432
✓
108
C
36
D
18

Answer

Correct option: B.

108

(B) 108
Explanation : Suppose, we fix 3 at unit place, then the total number possible arrangement for remaining numbers $=3$ !
Similarly, if we fix 4,5 and 6 at unit place, then in each case total possible arrangements are 3 !.
Hence, required number of arrangements in which sum of unit digits of all such numbers is
$
\begin{array}{l}
=3 \times 3!+4 \times 3!+5 \times 3!+6 \times 3! \\
=(3+4+5+6) \times 3! \\
=18 \times 6=108
\end{array}
$

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MCQ 61 Mark

In an examination there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answer correct is

A
11
B
12
C
27
✓
63

Answer

Correct option: D.

(D) 63
Explanation : There are three multiple choice questions, each has four possible answers. Therefore, the total number of possible answers will be $4 \times 4 \times 4=64$. Out of these possible answers only one will be correct and hence number of ways in which a student can fail to get correct answer is $64-1=63$

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MCQ 71 Mark

All the letters of the word 'EAMCOT' are arranged in different possible ways. The number of such arrangements in which no two vowels are adjacent to each other is

A
360
✓
144
C
72
D
54

Answer

Correct option: B.

144

(B) 144
Explanation : We note that, there are 3 consonants and 3 vowels, first let us arrange the consonants in a row. This can be done in ${ }^3 P_3=3!=6$ ways.
$
\times M \times C \times T \times
$
Now, no. two vowels are adjacent to each other if they put at places marked $\times$. The 3 vowels can fill up these 4 places in ${ }^4 P_3=4 \times 3 \times 2 \times 1=24$ ways. Hence, total number of arrangements $=6 \times 24=$ 144.

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MCQ 81 Mark

There are four bus routes between $A$ and $B$, and three bus routes between $B$ and $C$. A man can travel round-trip in number of ways by bus from $A$ to $C$ via $B$. If he does not want to use a bus route more than once, in how many ways can he make round trip ?

✓
72
B
144
C
14
D
19

Answer

Correct option: A.

(A) 72
Explanation : In the following figure, there are 4 bus routes from $A$ to $B$ and 3 routes from $B$ to $C$. Therefore, there are $4 \times 3=12$ ways to go from $A$ to C.

It is round trip so that the man will travel back from $C$ to $A$ via $B$. It is restricted the man cannot use same bus routes from $C$ to $B$ and $B$ to $A$ more than once. Thus, There are $2 \times 3=6$ routes for return journey. Therefore, the required number of ways $=12 \times 6=$ 72.

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