Fill in the blanks.

Question 11 Mark

Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done if atleast 2 are red is _________________ .

Answer

80 , because
The possible selections can be either 'two red and one other than red' or 'three red'.
$\begin{aligned} \text { So, required number of ways } & =\left({ }^5 C_2 \times{ }^7 C_1\right)+{ }^5 C_3 \\ & =(10 \times 7)+10 \\ & =80 .\end{aligned}$

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Question 21 Mark

Every body in a room shakes hand with everybody else. The total number of hand shakes is 66 . The total number of persons in the room is _________________ .

Answer

12, because
Let the total number of persons in a room be $n$.
Since, two persons make 1 hand-shake,
$\therefore$ The number of hands shakes $={ }^n C_2$
According to question,
$
\begin{aligned}
& & { }^n C_2 & =66 \\
\Rightarrow & & \frac{n!}{2!(n-2)!} & =66 \\
\Rightarrow & & \frac{n(n-1)(n-2)!}{2 \times 1 \times(n-2)!} & =66 \\
\Rightarrow & & \frac{n(n-1)}{2} & =66 \\
\Rightarrow & & n^2-n-132 & =0 \\
\Rightarrow & & n^2-12 n+11 n-132 & =0 \\
\Rightarrow & & (n-12)(n+11) & =0 \\
\Rightarrow & & n & =12, n=-11 \\
& \therefore & n & =12
\end{aligned}
$
[ $\because n \neq-11$ i.e., persons cannot be negative]

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Question 31 Mark

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is _________________

Answer

18, because
We know that, to form a parallelogram, we need a pair of lines from the set of 4 lines and another pair of lines from another set of 3 lines.
Therefore, required number of parallelogram
$
\begin{array}{l}
={ }^4 C_2 \times{ }^3 C_2 \\
=6 \times 3=18
\end{array}
$

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Question 41 Mark

${ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7= $ _________________

Answer

0, because
$
\begin{array}{l}
{ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7 \\
={ }^{15} C_{15-8}+{ }^{15} C_{15-9}-{ }^{15} C_6-{ }^{15} C_7 \quad\left[\because{ }^n C_r={ }^n C_{n-r}\right] \\
={ }^{15} C_7+{ }^{15} C_6-{ }^{15} C_6-{ }^{15} C_7 \\
=0 .
\end{array}
$

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Question 51 Mark

The number of signals that can be sent by 6 flags of different colours taking one or more at a time is _________________ .

Answer

1956, because
Number of signals using one flag $={ }^6 P_1=6$
Number of signals using two flags $={ }^6 P_2=30$
Number of signals using three flags $={ }^6 P_3=120$
Number of signals using four flags $={ }^6 P_4=360$
Number of signals using five flags $={ }^6 P_5=720$
Number of signals using six flags $={ }^6 P_6=720$
Therefore, the total number of signals using one or more flags at a time
$=6+30+120+360+720+720$ (using addition principle)
=1956

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Question 61 Mark

The number of six-digit numbers, all digits of which are odd is _________________

Answer

$(5)^6$, because
Out of the digits $0,1,2,3,4,5,6,7,8,9$; the odd digits are $1,3,5,7,9$.
Therefore, number of 6 digits numbers $=(5)^6$.

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Question 71 Mark

The total number of 9 digit numbers which have all different digits is _________________

Answer

$9 \times 9$ !, because
We have to form 9 digit numbers from the digits 0 , $1,2,3,4,5,6,7,8,9$ and we know that 0 cannot be put on the extreme left place.
So, first place from the left can be filled in 9 ways.
Now, for the remaining 8 places repetition is not allowed. So, the remaining 8 places can be filled in 9! ways.
Hence, total number of ways $=9 \times 9$ !

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Question 81 Mark

The number of possible outcomes when a coin is tossed 6 times is _________________

Answer

64, because
We know that, coin has two sides i.e., head $( H )$ and tail (T)
$\therefore$ When a coin is tossed 6 times, then the possible outcomes $=2^6=64$

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