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Binomial Theorem · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 21 questions.

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Question 13 Marks
If $^nC_p = ^nC_p Find ^{12}C_n$.
Answer
We have, If ${ }^n C_p={ }^n C_q=n$ Then $p+q=n$
Also, ${ }^n C_r=\frac{n!}{r!(n-r)!} \ldots$ (i)
$\Rightarrow{ }^n C_4={ }^n C_6 4+6=n \Rightarrow n=10$
Applying (i), ${ }^{12} \mathrm{C}_{10}=\frac{12!}{10!2!}=\frac{12 \times 11 \times 10!}{10!\times 2 \times 1}=\frac{12 \times 11}{2 \times 1}=66$
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Question 23 Marks
How many different boat parties of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?
Answer
We have, Total boy = 25 Total girls = 10 Party of 8 to be made 25 boy and 10 girls, $= {^\text{25}\text{C}}_{\text{5}},{^\text{10}\text{C}}_{\text{3}}$ $= {^\text{25}\text{C}}_{\text{5}}\times{^\text{10}\text{C}}_{\text{3}}$ Now, ${^\text{25}\text{C}}_{\text{5}}=\frac{\text{n}!}{\text{r}!(\text{n}-\text{r})!}$ $=\frac{25!}{5!20}\times\frac{10}{3!7!}$ $=\frac{25\times24\times23\times22\times21\times10\times9\times9\times8}{5\times4\times3\times2\times3\times2}$ $=6375600$
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Question 33 Marks
Find the number of diagonals of:
  1. A hexagon.
  2. A polygon of 16 sides.
Answer
  1. A hexagon:
A hexagon has 6 angular points. By joining any two angular points we get a line which is either a side or a diagonal.
Number of line $={^{6}{\text{C}}}_{\text{2}}=\frac{6!}{2!4!}$
$=\frac{6\times5}{2}=15$
Number of sides = 6
Number of diagonals = 15 - 6 = 9
  1. A polygon of 16 sides:
A polygonof 16 sides will have 16 angular points. By joining any 2 points we get a line which is either a side or a diagonal.
Number of lines $={^{16}{\text{C}}}_{\text{2}}=\frac{16!}{2!14!}$
$=\frac{16\times15}{2}=120$
Number of sides = 16
Number of diagonals = 120 - 16 = 104
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Question 43 Marks
How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
Answer
INVOLUTE Number of letter = 8 Wovels = I, O, U, E Consonents = N, V, L, T Number of ways to select 3 wovels $={^\text{4}}\text{C}_{\text{3}}$ Number of ways to select 2 Consonents $={^\text{4}}\text{C}_{\text{2}}$ Number of ways to arrange these five letters, $={^\text{4}}\text{C}_{\text{3}}\times​​{^\text{4}}\text{C}_{\text{2}}\times5!$ $=4\times6\times5\times4\times3\times2\times1$ $=2880$
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Question 53 Marks
If ${ }^n C_{10}={ }^n C_{12}$, Find ${ }^{23} C_n$.
Answer
We have, If ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{p}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{q}}=\mathrm{n}$
Then $\mathrm{p}+\mathrm{q}=\mathrm{n}$
Also, $\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{10}={ }^{\mathrm{n}} \mathrm{C}_{12} 10+12=\mathrm{n} \Rightarrow \mathrm{n}=22$
Applying (i), ${ }^{23} \mathrm{C}_{22}=\frac{23!}{22!1!}$
$=\frac{23 \times 22!}{22!}=23$
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Question 63 Marks
We wish to select 6 persons from 8 but if the person A is chosen, then B must be chosen. In how many ways can the selection be made?
Answer
We have, Total persons = 8 Selection to be made = 6 person. If A is chosen then B must be chosen. ⇒ A and B are chosen together Selection can be made in, $\Rightarrow {^{6}{\text{C}}}_{\text{4}}=\frac{6!}{4!2!}$ $=\frac{6\times5}{2}=15$ Also the number of selection in which A and B are not chosen are, $\Rightarrow {^{7}{\text{C}}}_{\text{6}}=\frac{7!}{6!1!}$ $=7$ Total number of ways in which selection is made = 15 + 7 = 22
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Question 73 Marks
Prove that: $^{4\text{n}}\text{C}_{2\text{n}}:^{2\text{n}}\text{C}_{\text{n}}=[1,3,5...(4\text{n}-1):[1,3,5...(2\text{n-1})^{2}].$
Answer
We have, $\Rightarrow \frac{\frac{4\text{n}!}{(2\text{n})!(2\text{n}!)}}{\frac{2\text{n}!}{\text{n}!\text{n}!}}$ $\Rightarrow \frac{(4\text{n}!)\times(\text{n}!^{2})}{(2\text{n}!)(2\text{n})\times(2\text{n})!^{2}}$ $\Rightarrow \frac{\big[1,2,3,4...(4\text{n}-1)(4\text{n})\big](\text{n}!^{2})}{(2\text{n})!\big[1,2,3,4....(2\text{n}-2)(2\text{n}-1)(2\text{n})\big]^{2}}$ $\Rightarrow \frac{\big[1,3,5...(4\text{n}-1)\big]\times\big[2,4,6...4\text{n}\big](\text{n}!^{2})}{(2\text{n})!\big[1,3,5....(2\text{n}-1)\big]^{2}\times\big[2,4,6....(2\text{n}-2)(2\text{n})\big]^{2}}$ $\Rightarrow \frac{\big[1,3,5.....(4\text{n}-1)\big]}{\big[1,3,5.....(4\text{n}-1)\big]^{2}}$ Hence proved.
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Question 83 Marks
A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions?
Answer
Total number of quation = 10 Quation in part A = 6 Quation in part B = 11Selecting to quation with at least 4 from each part A and B. Can from done in followin way.
${^{6}{\text{C}}}_{\text{4}}\times{^{7}{\text{C}}}_{\text{6}}+{^{6}{\text{C}}}_{\text{5}}\times{^{7}{\text{C}}}_{\text{5}}+{^{6}{\text{C}}}_{\text{6}}\times{^{7}{\text{C}}}_{\text{4}}$ $=2\Big({^{20}{\text{C}}}_{\text{6}}\times{^{20}{\text{C}}}_{\text{5}}\Big)$ $=\Big(\frac{6!}{4!2!}\times\frac{7!}{6!1!}\Big)+\Big(\frac{6!}{5!1!}\times\frac{7!}{5!2!}\Big)+\Big(\frac{1\times7!}{4!3!}\Big)$ $=\big(\frac{6\times5\times7}{2}\big)+\big(\frac{6\times7\times6}{2}\big)+\big(\frac{7\times6\times\times5}{3\times2}\big)$ $=(105)+(126)+(35)$ $=266$
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Question 93 Marks
Prove that the product of 2n consecutive negative integers is divisible by (2n!).
Answer
We have, Product by, $=\big[(2\text{n}+1)(2\text{n}+3)(2\text{n}+5)....(2\text{n}+\text{r})\big]$ $=\frac{(2\text{n})\big[(2\text{n+1})(2\text{n+3})....(2\text{n}+\text{r})\big]}{(2\text{n}!)}$ $=\frac{(2\text{n}+\text{r})!}{(2\text{n})!}$ Hence r = 2n $=\frac{(2\text{n}+2\text{n})!}{2\text{n}}$ $=\frac{(4\text{n})!}{(2\text{n})!}$ $=(2\text{n})!$
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Question 103 Marks
If ${^\text{2n}}\text{C}_{\text{3}}:{^\text{n}}\text{C}_{\text{2}}=44:3,$ find n.
Answer
We have, $\Rightarrow \frac{\frac{2\text{n}!}{(3\text{n})!(2\text{n-3}!)}}{\frac{\text{n}!}{\text{2}!(\text{n-2}!)}}$ $\Rightarrow \frac{2\text{n}!2!(\text{n}-2)!}{3!(2\text{n}-3)!\text{n}!}=\frac{44}{3}$ $\Rightarrow \frac{2\text{n}!}{3\text{n}~(\text{n-1})!(2\text{n}-3)!}=\frac{44}{3}$ $\Rightarrow 2\text{n}(2\text{n}-1)(2\text{n}-2)=44\text{n}(\text{n}-1)$ $\Rightarrow (2\text{n}-1)(\text{n}-1)=11(\text{n}-1)$ $\text{n}=6$
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Question 113 Marks
If ${^\text{n}}\text{C}_{12}={^\text{n}}\text{C}_{5},$ Find the value of n.
Answer
We have, ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}$ Hence, $\mathrm{n}=\mathrm{mr}=12$ and 5 Applying formula ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{p}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{q}}=\mathrm{n}$ Then $\mathrm{p}+\mathrm{q}=\mathrm{n}={ }^{\mathrm{n}} \mathrm{C}_{12}={ }^{\mathrm{n}} \mathrm{C}_5 12$ $+5=n \Rightarrow n=17$
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Question 123 Marks
In an examination, a question paper consists of 12 questions divided into two parts i.e. Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Answer
Here, part I has quations and part 11 has 7 quations. Student has to attempt 8 quations selecting at least 3 from each section. So, Number of ways to select at least 3 from each section and a total 8 quations. = (3 from part I an 5 from part II) or (4 from part I from part II) or (5 from part I and 3 from part II) $=\big({^\text{5}}\text{C}_{\text{3}}\times{^\text{7}}\text{C}_{\text{5}}\big)+\big({^\text{5}}\text{C}_{\text{4}}\times{^\text{7}}\text{C}_{\text{5}}\big)+\big({^\text{5}}\text{C}_{\text{3}}\times{^\text{7}}\text{C}_{\text{3}}\big)$ $=\Big(\frac{5\times4}{2\times1}\times\frac{7\times6}{2\times1}\Big)+\Big(5\times\frac{7\times6\times5}{3\times2}\Big)+\Big(1\times7\frac{5\times6\times5}{3\times2}\Big)$ $=210+170+30$ $=420$
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Question 133 Marks
For all positive integers n, show that ${^{2\text{n}}}\text{C}_{\text{n}}+{^{2\text{n}}}\text{C}_{\text{n}-1}=\frac{1}{2}\big({^{2\text{n+2}}}\text{C}_{\text{n}+1}\big).$
Answer
We have, $\text{L.H.S.}={^{2\text{n}}}\text{C}_{\text{n}}+{^{2\text{n}}}\text{C}_{\text{n}-1}$ $=\frac{2\text{n}!}{\text{n}!\text{n}!}+\frac{2\text{n}!}{(\text{n}-1)!(\text{n}-1)!}$ $=(2\text{n})!\Big[\frac{1}{\text{n}(\text{n}-1)!(\text{n})(\text{n}-1)!}+\frac{1}{(\text{n}-1)!(\text{n}-1)!}\Big]$ $=\frac{(2\text{n}!)}{(\text{n}-1)!(\text{n}-1)!}\Big[\frac{1+\text{n}^{2}}{\text{n}^{2}}\Big]\ ...(\text{i})$ ${^{2\text{n+2}}}\text{C}_{\text{n}+1}=\frac{(2\text{n}+2)!}{(\text{n+1}!)(\text{n}+1)!}$ $=\frac{(2\text{n}+2)(2\text{n}+1)(2\text{n}!)}{\text{n}(\text{n}+1)\text{n}(\text{n}-1)!}\ ...(\text{ii})$ $=\frac{(2\text{n}!)}{(\text{n}-1)!(\text{n}-1)!}\times\frac{(\text{n}+1)^{2}(\text{n}^{2})(\text{n}-1)!(\text{n}-1)!}{(2\text{n}-2)(2\text{n}+1)(2\text{n}!)}\times\Big(\frac{1+\text{n}^{2}}{\text{n}^{2}}\Big)$ $=\frac{(\text{n}+1)(\text{n}^{2}+1)}{(2\text{n}+1)}\times\frac{1}{2}$
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Question 143 Marks
A business man hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?
Answer
In One round table the business man can accommodate the guests in ways. In the second round table he can the guests in ways. Keeping one guest as fixed in the round table, the other 14 guests can be arrange in 14! ways. Keeping one guest as fixed in the second round tabie, the other 5 guests can be number of ways in which the guests can be arrange is $={^\text{21}}\text{C}_{\text{15}}\times{^\text{6}}\text{C}_{\text{6}}\times14!\times5!\ \text{ways}$
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Question 153 Marks
There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.
Answer
Number of point = 10 Number of collinear points = 4 Since 4 out of 10 points are collinear, so the number of liner will be, Number of liner ${^{10}{\text{C}}}_{\text{2}}=\big({^{4}{\text{C}}}_{\text{2}}-1\big)$ $={^{10}{\text{C}}}_{\text{2}}-{^{4}{\text{C}}}_{\text{2}}+1$ $=\frac{10!}{2!8!}-\frac{4!}{2!2!}+1$ $=\frac{10\times9}{2}-\frac{4\times3}{2}+1$ $=45-6+1$ $=40$
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Question 163 Marks
Evaluate the following: $^{n+1}C_n$
Answer
We have, $=\frac{(\text{n}+1)!}{(\text{n!})(\text{n+1-n})!}$ $=\frac{(\text{n+1})\times\text{n}!}{\text{n!}\times1!}$ $=\text{n}+1$
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Question 173 Marks
If ${^\text{15}}\text{C}_{\text{r}}:{^\text{15}}\text{C}_{\text{r-1}},=11:5,$ Find r.
Answer
We have, $\frac{{^\text{15}}\text{C}_{\text{r}}}{{^\text{15}}\text{C}_{\text{r}-1}}=\frac{11}{5}$ $\Rightarrow \frac{15-\text{r}+1}{\text{r}}=\frac{11}{5}$ $\Rightarrow 75-5\text{r}+5=11\text{r}$ $\Rightarrow\ 16\text{r}=80$ $\Rightarrow \text{r}=5$
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Question 183 Marks
If ${^\text{15}}\text{C}_{3\text{r}}={^\text{15}}\text{C}_{\text{r+3}},$ Find $r$.
Answer
We have, If ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{p}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{q}}=\mathrm{n}$
Then $\mathrm{p}+\mathrm{q}=\mathrm{n}$
Also, ${ }^{15} \mathrm{C}_{3 \mathrm{r}}={ }^{15} \mathrm{C}_{\mathrm{r}+3}, $
$\Rightarrow 3 \mathrm{r}+\mathrm{r}+3=154 \mathrm{r}+3=154 \mathrm{r}=15-34 \mathrm{r}=12 \mathrm{r}$ $=3$
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Question 193 Marks
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer
We have, There are 5 boys and 4 girls. The team consists of 3 boys and 3 girls. Number of ways to from the learm $={^{5}{\text{C}}}_{\text{3}}\times{^{4}{\text{C}}}_{\text{3}}$ $=\frac{5!}{3!2!}\times\frac{4!}{3!}$ $=\frac{5\times4}{2}\times4$ $=40$
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Question 203 Marks
If ${ }^{18} C_x={ }^{18} C_{x+2}$, Find $x$.
Answer
We have, If ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{p}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{q}}=\mathrm{n}$
Then $\mathrm{p}+\mathrm{q}=\mathrm{n}$
Also, ${ }^{18} \mathrm{C}_{\mathrm{x}}={ }^{18} \mathrm{C}_{\mathrm{x}+2}$ $\Rightarrow x+x+2=182 x+2=182 x=18-22 x=16 x=8$
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Question 213 Marks
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
  1. Exactly 3 girls?
  2. At least 3 girls?
  3. At most 3 girls?
Answer
  1. The committee consists of exaclty 3 girls.
we have to select 4 boys from 9 boys.
This can be done in ways nd 3 girls out of 4 girls can br selected in ways.
The required number ways $={^\text{9}}\text{C}_{\text{4}}\times{^\text{4}}\text{C}_{\text{3}}$
$=\frac{9\times8\times7\times6}{4\times3\times2\times1}\times4$
$=504$
  1. At least 3 girls are there.
There are 3 or more than 3 or 4 girls.
  1. 3 girls and 4 boys $={^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{3}}$
  2. 4 girls and 3 boys $={^\text{4}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{3}}$
The required number ways $={^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{4}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}$
$=504+84$
$=588$
  1. For at most 3 girls theres are 3, 2, 1.
  1. 0 girls and 7 boys $={^\text{4}}\text{C}_{\text{0}}\times{^\text{9}}\text{C}_{\text{7}}$
  2. 1 girls and 7 boys $={^\text{4}}\text{C}_{\text{1}}\times{^\text{9}}\text{C}_{\text{6}}$
  3. 2 girls and 7 boys $={^\text{4}}\text{C}_{\text{2}}\times{^\text{9}}\text{C}_{\text{5}}$
  4. 3 girls and 7 boys $={^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{4}}$
Total number of required ways
$={^\text{4}}\text{C}_{\text{0}}\times{^\text{9}}\text{C}_{\text{7}}+{^\text{4}}\text{C}_{\text{1}}\times{^\text{9}}\text{C}_{\text{6}}+{^\text{9}}\text{C}_{\text{2}}\times{^\text{9}}\text{C}_{\text{5}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{4}}$
$=0\times\frac{9\times8}{2}+4\times\frac{9\times8\times7}{3\times2}+\frac{4\times3}{2}\times\frac{9\times8\times7\times6}{4\times3\times2}+504$
$=36+48\times7+18\times42+504$
$=1630$
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