MATHS 5 Marks Questions

1. Since, the team does not indude any girl therefore, only boys are to be selected.

5 boys out of 7 boys can be selected in ways.

$\Rightarrow {^{7}{\text{C}}}_{\text{5}}=\frac{7!}{5!2!}$

$=\frac{6\times7}{2}=21$

2. Since, the least ne boy and girls are to be there in every team. The team consist of:

1. 1 boy and 4 girls ${^{7}{\text{C}}}_{\text{1}}\times{^{4}{\text{C}}}_{\text{4}}$ 

2. 2 boy and 3 girls ${^{7}{\text{C}}}_{\text{2}}\times{^{4}{\text{C}}}_{\text{3}}$

3. 3 boy and 2 girls ${^{7}{\text{C}}}_{\text{3}}\times{^{4}{\text{C}}}_{\text{2}}$

4. 4 boy and 1 girls ${^{7}{\text{C}}}_{\text{4}}\times{^{4}{\text{C}}}_{\text{1}}$

The required number of ways,

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

$=7+84+210+140$

$=441$

3. Since, the tearm has to consist of at least 3 girls, the tearm can consist of,

1. 3 girls and 2 boys ${^{7}{\text{C}}}_{\text{2}}\times{^{4}{\text{C}}}_{\text{3}}$

2. 4 girls and 1 boys ${^{4}{\text{C}}}_{\text{4}}\times{^{7}{\text{C}}}_{\text{1}}$

The required number of ways,

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

$=84+7$

$=91$

$\text{n}\times {{^\text{n-1}}\text{C}_{\text{r-1}}}$

$=\text{n}\times\frac{(\text{n}-1)!}{(\text{r}-1)!(\text{n}-\text{r})!(\text{r}-1)!}$

$=\frac{{\text{n}!\times(\text{n}-\text{r}+1)}}{(\text{r}-1)!(\text{n}-\text{r})!(\text{n}-\text{r}+1)}$

Multiplying numerator by,

$=\frac{(\text{n}-\text{r}+1)\times\text{n}!}{(\text{r}-1)!(\text{n}-\text{r}+1)!}$

$=(\text{n}-\text{r}+1)^{\text{n}}\text{C}_{\text{r}-1}$

Hence proved

1. 18 families out of 52 and 2 families out of 35.
2. 19 families out of 52 and 1 families out of 35.
3. 20 families out of 52.

1. All different $= {^\text{8}}\text{C}_{\text{4}}\times4!={^\text{8}}\text{C}_{\text{4}}=\frac{8!}{4!}$

$=8\times7\times6\times5$

$=56\times30$

$=1680$

2. 2 distinct and 2 alike $= {^\text{3}}\text{C}_{\text{1}}\times{^\text{7}}\text{C}_{\text{2}}=\frac{3\times7\times6}{2}$ 

$=63\times\frac{4!}{2!}$

$=378$

3. 2 alike of one and 2 distinct letter $= {^\text{3}}\text{C}_{\text{2}}\times\frac{4!}{2!2!}$

$=3\times6=18$

3. 3 alike of one and 2 distinct letter $= {^\text{3}}\text{C}_{\text{1}}\times{^\text{7}}\text{C}_{\text{2}}=\frac{3\times7\times6}{2}$

$=378$

Total number of ways in which 4 letter are formes = 1680 + 378 + 18 + 378

= 2454 ways

1. There is no restriction.

$\Rightarrow {^\text{10}}\text{C}_{4}=\frac{10!}{4!6!}$

$=\frac{10\times9\times8\times7}{4\times3\times2}$

$=210$

2. Two particular books are always selected

These the total books = 10 - 2 = 8

So out of remaining 8 books selection and 2 books can be done in way,

$\Rightarrow {^\text{8}}\text{C}_{2}=\frac{8!}{2!6!}$

$=\frac{8\times7}{2\times1}=28$

3. Two particular books are never selected

These the total books = 10 - 2 = 8

So out of remaining 8 books selection and 2 books can be done in way,

$\Rightarrow {^\text{8}}\text{C}_{4}=\frac{8!}{4!4!}$

$=\frac{8\times7\times6\times5}{4\times3\times2}=70$ 

1. Include 2 particular players:

Now we have to select 9 more out of remaining 14

Required number of ways $={^\text{14}}\text{C}_{9}$

$=\frac{14!}{9!5!}$

$=\frac{14\times13\times12\times11\times10}{5\times4\times3\times2}=2002$

2. Exclude 2 particular players:

Now we have to select 11 players more out of remaining 14 players $={^\text{14}}\text{C}_{11}$

$=\frac{14!}{11!3!}$

$=\frac{14\times13\times12}{3\times2}=364$

1 man can be selected from 2 men in ways and 2 wimen can be selected fron 3 ways,

The required number of committees

1. To include exactly one officer.

This can be done is ${^\text{4}}\text{C}_{1}\times{^\text{8}}\text{C}_{5}$

$=\frac{4!}{1!3!}\times\frac{8!}{5!3!}$

$=\frac{4\times8\times7\times6}{3\times2}=224$

2. To include at least one officer

This can be done is

 ${^\text{4}}\text{C}_{1}\times{^\text{8}}\text{C}_{5}+{^\text{4}}\text{C}_{2}\times{^\text{8}}\text{C}_{4}+{^\text{4}}\text{C}_{1}\times{^\text{8}}\text{C}_{3}+{^\text{4}}\text{C}_{4}\times{^\text{8}}\text{C}_{2}$

$=\frac{4\times8!}{5!3!}+\frac{4!}{2!2!}+\frac{8!}{4!4!}+\frac{4!}{3!1!}\times\frac{8!}{3!5!}+\frac{1\times8!}{2!6!}$

$=\big(\frac{4\times8\times7\times6}{3\times2}\big)+\big(\frac{4\times3\times8\times7\times6\times5}{2\times4\times3\times2}\big)+\big(\frac{4\times8\times7\times6}{3\times2}\big)+\big(\frac{8\times7}{2\times1}\big)$

$=(4\times8\times7)+(4\times3\times7\times5)+(4\times8\times7)+(4\times7)$

$=224+420+224+28$

$=896$

1. A particular professor is included:

Committee is ${^\text{9}}\text{C}_{1}\times{^\text{20}}\text{C}_{3}$

$=\frac{9!}{8!}$

$=\frac{20}{3!\times17!}=\frac{9\times20\times19\times18}{3\times2}=10260$

2. A particular student is included:

 Committee is ${^\text{10}}\text{C}_{2}\times{^\text{19}}\text{C}_{2}$

$=\frac{10!}{2!8!}\times\frac{19!}{2!17!}$

$=\frac{10\times9\times19\times18}{2\times2\times1}=7695$

3.  A particular student is excluded:

Committee is ${^\text{10}}\text{C}_{2}\times{^\text{19}}\text{C}_{3}$

$=\frac{10!}{2\times8!}\times\frac{19!}{3!16!}$

$=\frac{10\times9\times19\times18\times17}{2\times3\times2}=43605$

1. Case I: There are 6 different letters is which all the are district to selected.

Number of ways to select therefour $={^\text{6}}\text{C}_{\text{4}}$

$=15$

Case II: Two same and two district letter are selected there are more than. letter.

Number of ways to select therefour $={^\text{3}}\text{C}_{\text{1}}\times{^\text{5}}\text{C}_{\text{2}}$

$=3\times10$

$=30$

Case III: Two alike of one kind two alike of other kind.

There are 3 pairs of letters is the more than one letters.

$={^\text{3}}\text{C}_{\text{2}}$

$=3$

Case IV: 3 alike and one different.

Number of ways to select these letters

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

$=5$

Number of ways to select four letters

= 15 + 30 + 3 + 5

= 53

2. For,

Case I: Number of arrangements of four letters all distiect $={^\text{6}}\text{C}_{\text{4}}\times4!$

$=15\times24$

$=360$

Case II: Number of arrangents to for letter two alike of kind and two of other,

$={^\text{3}}\text{C}_{\text{2}}\times\frac{4!}{2!2!}$

$=3\times10\times12$

$=360$

Case III: Number of arrangements to four letters 3 alike and 1 kind,

 $={^\text{3}}\text{C}_{\text{2}}\times\frac{4!}{2!2!}$

$=3\times6$

$=18$

Case IV: Number of arrangements to four letters 3 alike and 1 kind,

$=1\times{^\text{5}}\text{C}_{\text{1}}\times\frac{4!}{3!1!}$

$=20$

Total number of ways = 360 + 360 + 18 + 20

Required number of arrangements = 758

1. Number of arrangement of 4 letters.

Selected from these $={^\text{6}}\text{C}_{\text{4}}\times4!$

$=15\times24$

$=360$

2. Two ailke and with more than one

So, one pair from these and 2 from letters from rest 5 letters.

Number of ways to arrange therefour

$={^\text{2}}\text{C}_{\text{1}}\times​​{^\text{5}}\text{C}_{\text{2}}\times\frac{4!}{2!}$

$=2\times10\times12$

$=240$

3. Two ailke and with more than

Number of ways to arrange therefour

$={^\text{2}}\text{C}_{\text{2}}\times​​{^\text{5}}\text{C}_{\text{2}}\times\frac{4!}{2!2!}$

$=6$

4. There alike and one different number of ways to the therefour

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

$=5\times\frac{4!}{3!1!}$

$=20$

Required number of ways = 240 + 360 + 6 + 20

Required number of ways = 626

Number of sides = 10

Number of diagonals = 45 - 10 = 35

Also, by joining 3 angular points a triangle in formed