MATHS M.C.Q (1 Marks)

2. $2^{8}-2$

Solution:

$({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{3}}+{^\text{7}}\text{C}_{\text{4}})+......$

$=1+2\times{^\text{7}}\text{C}_{\text{1}}+2\times{^\text{7}}\text{C}_{\text{2}}+2\times{^\text{7}}\text{C}_{\text{3}}+2\times{^\text{7}}\text{C}_{\text{4}}+2\times{^\text{7}}\text{C}_{\text{5}}..$

$=2+2^{2}({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})​$

$=2+2^{2}(7+\frac{7}{2}\times6+\frac{7}{3}\times\frac{6}{2}\times5)$

$=2+252$

$=254$

$=2^{8}-2$

2. 56

Solution:

$\text{r}+\text{r}+4=20$

$\Rightarrow 2\text{r}+4=20$

$\Rightarrow 2\text{r}=16$

$\Rightarrow \text{r}=8$

Now,

${^\text{r}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$

$\therefore\ {^\text{8}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$

$\therefore\ {^\text{8}}\text{C}_{\text{3}}=\frac{8!}{3!5!}$

$=\frac{8\times7\times6}{3\times2\times1}=56$

1. 60

Solution:

Three persons can take 5 seats in ⁵C₃ ways. Moreover 3 persons can sit in 3! ways.

Required number of ways ${^\text{5}}\text{C}_{\text{3}}\times3!$

$=10\times6=60$

3. 3

Solution:

$3\text{r}+\text{r}+3=15$

$\Rightarrow 4\text{r}+3=15$

$\Rightarrow 4\text{r}=12$

$\Rightarrow \text{r}=3$

2. 816

Solution:

$\text{r}+\text{r}-10=20$

$\Rightarrow 2\text{r}-10=20$

$\Rightarrow 2\text{r}=30$

$\Rightarrow \text{r}=15$

Now,

${^\text{18}}\text{C}_{\text{r}}={^\text{18}}\text{C}_{\text{15}}$

$\therefore\ {^\text{18}}\text{C}_{\text{15}}={^\text{18}}\text{C}_{\text{3}}$

$\therefore\ {^\text{18}}\text{C}_{\text{3}}=\frac{18}{3}\times\frac{17}{2}\times16$

$=816$

2. 3

Solution:

$\text{a}^{2}-\text{a}=2+4$

$\Rightarrow \text{a}^{2}-\text{a}-6=0$

$\Rightarrow \text{a}^{2}-3\text{a}+2\text{a}-6=0$

$\Rightarrow \text{a}(\text{a}-3)+2(\text{a}-3)=0$

$\Rightarrow (\text{a}+2)(\text{a}-3)=0$

$\Rightarrow \text{a}=-2,\text{a}=3$

$\text{a}=3$

1. 12

Solution:

$\text{r}-6+3\text{r}+1=43$

$\Rightarrow 4\text{r}-5=43$

$\Rightarrow 4\text{r}=48$

$\Rightarrow \text{r}=12$

2. 78

Solution:

4 out of 13 players are bowlers.

In other words, 9 players are not bowlers.

A team of 11 is to be selected so as to include at least 2 bowlers.

Number of ways $={^\text{4}}\text{C}_{\text{2}}\times{^\text{9}}\text{C}_{\text{9}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{8}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}$

$=6+36+36$

$=78$

2. 31

Solution:

${^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{3}}+{^\text{5}}\text{C}_{\text{4}}+{^\text{5}}\text{C}_{\text{5}}$

$={^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{5}}$

$=2\times{^\text{5}}\text{C}_{\text{1}}+2\times{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{5}}$

$= 2\times5+2\times\frac{5!}{2!3!}+1$

$=10+20+1$

$=31$

4. 120

Solution:

Number of committes that can be formed $={^\text{6}}\text{C}_{\text{3}}\times{^\text{4}}\text{C}_{\text{2}}$

$=\frac{6!}{3!3!}\times\frac{4!}{2!2!}$

$=\frac{6\times5\times4}{3\times2}\times\frac{4\times3}{2}$

$=120$

1. 246

Solution:

Required number of ways $={^\text{4}}\text{C}_{\text{1}}\times{^\text{6}}\text{C}_{\text{4}}+{^\text{4}}\text{C}_{\text{2}}\times{^\text{6}}\text{C}_{\text{3}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{6}}\text{C}_{\text{2}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{6}}\text{C}_{\text{1}}$

$=60+120+60+6$

$=246$

2. 140

Solution:

Suppose there are two friends, A and B, who do not attend the party together.

If both of them do not attend the party, then the number of ways of selecting 6 guests $={^\text{8}}\text{C}_{\text{6}}=28$

If one of them attends the party, then the number of ways of selecting 6 guests $=2.{^\text{8}}\text{C}_{\text{5}}=112$

Total number of ways = 112 + 28 = 140

2. 40

Solution:

Number of straight lines formed by joining the 10 points if we take 2 points at a time $={^\text{10}}\text{C}_{\text{2}}=\frac{10}{2}\times\frac{9}{1}=45$

Number of straight lines formed by joining the 4 points if we take 2 points at a time $={^\text{4}}\text{C}_{\text{2}}=\frac{4}{2}\times\frac{3}{1}=6$

But, 4 collinear points, when joined in pairs, give only one line.

Required number of straight lines = 14 - 6 + 1 = 40

3. 2m = n(n - 1)

Solution:

${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}}$

$\Rightarrow \frac{\text{m!}}{1!(\text{m}-1)!}=\frac{\text{n!}}{2!(\text{n}-2)!}$

$\Rightarrow \frac{\text{m}(\text{m}-1)!}{(\text{m}-1)!}=\frac{\text{n}(\text{n}-1)(\text{n}-2)!}{2!(\text{n}-2)!}$

$\Rightarrow2\text{m}=\text{n}(\text{n}-1)$

3. 5

Solution:

${^\text{n+1}}\text{C}_{\text{3}}=2\times{^\text{n}}\text{C}_{\text{2}}$

$\Rightarrow \frac{(\text{n}+1)!}{3!(\text{n-2})!}=2\times\frac{\text{n}!}{2!(\text{n}-1)!}$

$\Rightarrow \text{n+1}=6$

$\Rightarrow \text{n}=5$

3. 64

Solution:

Number of straight lines joining 12 points if we take 2 points at a time $={^\text{12}}\text{C}_{\text{2}}$

$=\frac{12!}{2!10!}=66$

Number of straight lines joining 3 points if we take 2 points at a time $={^\text{3}}\text{C}_{\text{2}}=3$

Required number of straight lines = 66 - 3 + 1 = 64

1. 20

Solution:

An octagon has 8 vertices.

The number of diagonals of a polygon is given by $\frac{\text{n}(\text{n}-3)}{2}$

Number of diagonals of an octagon $=\frac{\text{8}(\text{8}-3)}{2}=20$

2. 120

Solution:

If set S has n elements, then C (n, k)C n, k is the number of ways of choosing k elements from S.

Thus, the number of subsets of SS of all possible values is given by,

$\text{C}(\text{n},0)+\text{C}(\text{n},1)+\text{C}(\text{n},3)+.....+\text{C}(\text{n},\text{n})=2^\text{n}$

Comparing the given equation with the above equation:

$2^\text{n}=256$

$\Rightarrow 2^\text{n}=2^{8}$

$\Rightarrow \text{n}=8$

$\therefore {^\text{2n}}\text{C}_{\text{2}}={^\text{16}}\text{C}_{\text{2}}$

$\Rightarrow {^\text{16}}\text{C}_{\text{2}}=\frac{16!}{2!4!}=\frac{16\times15}{2}=120$​​​​​​​

3. ${^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}$

Solution:

A host lady can invite 8 out of 12 people in ways. Two out of these 12 people do not want to attend the party together.

Number of ways $={^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}.$

1. 10

Solution:

$\text{r}+\text{1}+\text{r}-1=20$ $[\therefore \ ^\text{n}\text{C}_{\text{x}} = \ ^\text{n}\text{C}_{\text{y}} \Rightarrow \text{n} = \text{x} + \text{y} \text{ or } \text{x = y}]$

$\Rightarrow 2\text{r}=20$

$\Rightarrow \text{r}=10$

4. $\text{r}+1$

Solution:

We have,

${^\text{n}}\text{C}_{\text{r}}+{^\text{n}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}}$

$\Rightarrow {^\text{n+1}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}}$

$\Rightarrow \text{r}+1=\text{x}$

${^\text{n}}\text{C}_{\text{x}}={^\text{n}}\text{C}_{\text{y}}$

$\Rightarrow \text{n}=\text{x}+\text{y},\text{x}=\text{y}$

3. 264

Solution:

Among 14 players, 5 are bowlers.

A team of 11 players has to be selected such that at least 4 bowlers are included in the team.

Required number of ways $={^\text{5}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}+{^\text{5}}\text{C}_{\text{5}}\times{^\text{9}}\text{C}_{\text{6}}$

$=180+84$

$=264$

4. 18

Solution:

A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines.

Two parallel lines from the set of four parallel lines can be chosen in ${^\text{4}}\text{C}_{\text{2}}$ ways and two parallel lines from the set of 3 parallel lines can be chosen in ${^\text{3}}\text{C}_{\text{2}}$ ways.

Number of parallelograms that can be formed $={^\text{4}}\text{C}_{\text{2}}\times{^\text{3}}\text{C}_{\text{2}}$ 

$=\frac{4!}{2!2!}\times\frac{3!}{2!1!}$

$=6\times3=18$

3. 7200

Solution:

2 out of 4 vowels can be chosen in ⁴C₂ ways and 3 out of 5 consonants can be chosen in ⁵C₃ ways.

Thus, there are $({^\text{4}}\text{C}_{\text{2}}\times{^\text{5}}\text{C}_{\text{3}})$ groups, each containing 2 vowels and 3 consonants.

Each group contains 5 letters that can be arranged in 5! ways.

Required number of words $=({^\text{4}}\text{C}_{\text{2}}\times{^\text{5}}\text{C}_{\text{3}})\times5!$

$=60\times120=7200$

2. 156

Solution:

We need at least three points to draw a circle that passes through them.
Now, number of circles formed out of 11 points by taking three points at a time $={^\text{11}}\text{C}_{\text{3}}=165$

Number of circles formed out of 5 points by taking three points at a time $={^\text{5}}\text{C}_{\text{3}}=10$

It is given that 5 points lie on one circle.

Required number of circles = 165 - 10 + 1 = 156.

1. 20

Solution:

$\text{n}=12+8=20$

1. 231

Solution:

${^\text{n}}\text{C}_{\text{12}}={^\text{n}}\text{C}_{\text{8}}$

$\Rightarrow \text{n}=12+8=20$

${^\text{n}}\text{C}_{\text{x}}={^\text{n}}\text{C}_{\text{y}}$

$\Rightarrow \text{n}=\text{x}+\text{y}, \text{x}=\text{y}$

Now,

${^\text{22}}\text{C}_{\text{n}}={^\text{22}}\text{C}_{\text{20}}$

$=\frac{22}{2}\times\frac{21}{1}$

$=231$