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Maths 5 Marks Questions

Combinations · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 40 questions.

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Question 15 Marks
In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected?
Answer
We have,
Total men = 6
Total women = 4
Total person in committee = 5
This can be done in,
${^{4}{\text{C}}}_{\text{1}}\times{^{6}{\text{C}}}_{\text{4}}+{^{4}{\text{C}}}_{\text{2}}\times{^{6}{\text{C}}}_{\text{3}}+{^{4}{\text{C}}}_{\text{3}}+{^{6}{\text{C}}}_{\text{2}}+{^{4}{\text{C}}}_{\text{4}}\times{^{6}{\text{C}}}_{\text{1}}$
$=\Big(\frac{4\times6!}{4!\times2!}\Big)+\Big(\frac{4!}{2!2!}\times\frac{6!}{313!}\Big)+\Big(\frac{4!}{3!1!}\times\frac{6!}{2!4!}\Big)+(1\times6)$
$=\Big(\frac{4\times6\times5}{2}\Big)+\Big(\frac{4\times3}{2}\times\frac{6\times5\times4}{3\times2}\Big)+\Big(\frac{4\times6\times5}{2}\big)+(6)$
$=(60)+(120)+(60)+(6)$
$=246$
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Question 25 Marks
In a village, there are $87$ families of which $52$ families have at most $2$ children. In a rural development programme, $20$ families are to be helped chosen for assistance of which at least $18$ families must have at most $2$ children. In how many ways can the choice be made?
Answer
We have,
$52$ families have at most $2$ childern, while $35$ families have $2$ children.
The selection of $20$ families of which at least $18$ families must have $2$ children can be made as under.
  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 $52.$
Therefore the number of ways are $={^{52}{\text{C}}}_{\text{35}}\times{^{35}{\text{C}}}_{\text{2}}+{^{52}{\text{C}}}_{\text{19}}\times{^{35}{\text{C}}}_{\text{19}}+{^{35}{\text{C}}}_{\text{1}}+{^{52}{\text{C}}}_{\text{20}}+{^{35}{\text{C}}}_{\text{0}}$
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Question 35 Marks
Find the number of combinations and permutations of $4$ letters taken from the word $\text{'EXAMINATION'.}$
Answer
The word $\text{EXAMINATION}$ has letter where $\text{A, I, N}$ repeat twice.
The total number of letter $= 11$
The number of ways of selecting $4$ letters.
$\Rightarrow {^\text{11}}\text{C}_{\text{4}}=\frac{11!}{4!7!}=\frac{11\times10\times9\times8}{4\times3\times2}$
$=330$
The number of arranging of selecting $4$ letters.
  1. All different $= {^\text{8}}\text{C}_{\text{4}}\times4!={^\text{8}}\text{C}_{\text{4}}=\frac{8!}{4!}$
$=8\times7\times6\times5$
$=56\times30$
$=1680$
  1. $2$ distinct and $2$ alike $= {^\text{3}}\text{C}_{\text{1}}\times{^\text{7}}\text{C}_{\text{2}}=\frac{3\times7\times6}{2}$
$=63\times\frac{4!}{2!}$
$=378$
  1. $2$ alike of one and $2$ distinct letter $= {^\text{3}}\text{C}_{\text{2}}\times\frac{4!}{2!2!}$
$=3\times6=18$
  1. $3$ 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
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Question 45 Marks
If ${^\text{28}}\text{C}_{2\text{r}}:{^\text{24}}\text{C}_{\text{2r-4}},=225:11,$ Find r.
Answer
We have,
$\Rightarrow \frac{\frac{28!}{(2\text{r})!(28-2\text{r}!)}}{\frac{24!}{(2\text{r}-4)!(24-(2\text{r}-4))!}}$
$\Rightarrow \frac{28\times27\times26\times24!(2\text{r}4)!(28-2\text{r})}{(2\text{r})!(28-2\text{r})!24!}=\frac{225}{11}$
$\Rightarrow \frac{28\times27\times25}{2\text{r}\times(2\text{r}-1)\times(2\text{r}-2)(2\text{r}-3)}=2\text{r}(2\text{r-1})(2\text{r}-2)(2\text{r}-3)$
$\Rightarrow 11\times12\times13\times14=2\text{r}(2\text{r}-1)(2\text{r}-2)(2\text{r}-3)$
$\Rightarrow \text{r}=7$
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Question 55 Marks
How many different selections of $4$ books can be made from $10$ different books, if
  1. There is no restriction.
  2. Two particular books are always selected.
  3. Two particular books are never selected?
Answer
Total number of books $= 10$
Total number to be selected $= 4$
  1. There is no restriction.
$\Rightarrow {^\text{10}}\text{C}_{4}=\frac{10!}{4!6!}$
$=\frac{10\times9\times8\times7}{4\times3\times2}$
$=210$
  1. 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,
$\Rightarrow {^\text{8}}\text{C}_{2}=\frac{8!}{2!6!}$
$=\frac{8\times7}{2\times1}=28$
  1. 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,
$\Rightarrow {^\text{8}}\text{C}_{4}=\frac{8!}{4!4!}$
$=\frac{8\times7\times6\times5}{4\times3\times2}=70$
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Question 65 Marks
If $\alpha={^\text{m}}\text{C}_{2},$ then find the value of ${^{\alpha}}\text{C}_{2}.$
Answer
We have,
$\alpha={^\text{m}}\text{C}_{2}=\frac{\text{m}(\text{m}-1)}{2}$
${^{\alpha}}\text{C}_{2}=\frac{\alpha(\alpha-1)}{2}$
$=\frac{\Big(\frac{\text{m}(\text{m}-1)}{2}\Big)\Big(\frac{\text{m}(\text{m}-1)}{2}-1\Big)}{2}$
$=\frac{\text{m}(\text{m}-1)(\text{m}^{2}-\text{m}-2)}{2\times2\times2}$
$=\frac{\text{m}(\text{m}-1)(\text{m}+1)(\text{m}-2)}{8}$
$=\frac{\text{m}(\text{m}-1)(\text{m}+1)(\text{m}-2)}{4\times2}$
Multiplying with 3, numerator and denominator to make 4.
$=\frac{\text{m}(\text{m}+1)\text{m}(\text{m}-1)(\text{m}-2)}{4.3.2.1}$
$=\frac{3(\text{m}+1)\text{m}(\text{m}-1)(\text{m}-2)}{4!}$
$=3.{^{\text{m+1}}}\text{C}_{4}$
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Question 75 Marks
From $4$ officers and $8$ jawans in how many ways can $6$ be chosen:
  1. To include exactly one officer.
  2. To include at least one officer?
Answer
Total number of officer $= 4$
Total number of jawans $= 8$
Total number of selection to be made $= 6$
  1. To include exactly one officer.
This can be done is ${^\text{4}}\text{C}_{1}\times{^\text{8}}\text{C}_{5}$
$=\frac{4!}{1!3!}\times\frac{8!}{5!3!}$
$=\frac{4\times8\times7\times6}{3\times2}=224$
  1. To include at least one officer
This can be done is
${^\text{4}}\text{C}_{1}\times{^\text{8}}\text{C}_{5}+{^\text{4}}\text{C}_{2}\times{^\text{8}}\text{C}_{4}+{^\text{4}}\text{C}_{1}\times{^\text{8}}\text{C}_{3}+{^\text{4}}\text{C}_{4}\times{^\text{8}}\text{C}_{2}$
$=\frac{4\times8!}{5!3!}+\frac{4!}{2!2!}+\frac{8!}{4!4!}+\frac{4!}{3!1!}\times\frac{8!}{3!5!}+\frac{1\times8!}{2!6!}$
$=\big(\frac{4\times8\times7\times6}{3\times2}\big)+\big(\frac{4\times3\times8\times7\times6\times5}{2\times4\times3\times2}\big)+\big(\frac{4\times8\times7\times6}{3\times2}\big)+\big(\frac{8\times7}{2\times1}\big)$
$=(4\times8\times7)+(4\times3\times7\times5)+(4\times8\times7)+(4\times7)$
$=224+420+224+28$
$=896$
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Question 85 Marks
From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?
Answer
Total number of boys = 12
Total number of girls = 10
Total number of girls for the competition = 10 + 2 = 12
Total student chosen for competition = 10 - 2
Selection can br made in,
$={^\text{12}}\text{C}_{4}\times{^\text{8}}\text{C}_{4}+{^\text{12}}\text{C}_{5}\times{^\text{8}}\text{C}_{3}+{^\text{12}}\text{C}_{6}\times{^\text{8}}\text{C}_{2}$
$=\frac{12!}{4!8!}\times\frac{8!}{4!4!}+\frac{12!}{5!7!}\times\frac{8!}{3!5!}+\frac{12!}{6!6!}\times\frac{8!}{2!6!}$
$=\Big(\frac{12\times11\times10\times9\times8\times7\times6\times5}{4\times3\times2\times4\times3\times2}\Big)+\Big(\frac{12\times11\times10\times9\times8\times8\times7\times6}{5\times4\times3\times2\times3\times2}\Big)$
$=55440+44352+181104$
$=280896$
Total number of ways
$=857770-280896=104874$
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Question 95 Marks
Evaluate the following:
$\sum\limits^5_\text{r=1}\ ^5\text{C}_{\text{r}}$
Answer
We have,
$= { ^5}\text{C}_{\text{1}}+{ ^5}\text{C}_{\text{2}}+{ ^5}\text{C}_{\text{3}}+{^5}\text{C}_{\text{4}}+{ ^5}\text{C}_{\text{5}}$
$=\frac{5!}{1!4!}+\frac{5!}{2!3!}+\frac{5!}{3!2!}+\frac{5!}{4!1!}+\frac{5!}{5!0!}$
$=5+\frac{5\times4}{2}+\frac{5\times4}{2}+5+1$
$=5+10+10+5+1$
$=31$
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Question 105 Marks
If $^{24}C_x =\ ^{24}C_{2x+3},$ Find x.
Answer
We have,
If $^nC_p =\ ^nC_q = n$
Then $p + q = n$
Also,
$x + 2x + 3 = 24$
$3x = 21$
$x = 7$
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Question 115 Marks
Let r and n be positive integers such that 1 < r < n. Then prove the following:
$\frac{{{^\text{n}}\text{C}_{\text{r}}}}{{^\text{n-1}}\text{C}_{\text{r}-1}}=\frac{\text{n}}{\text{r}}$
Answer
We have,
${{^\text{n}}\text{C}_{\text{r}}}=\frac{\text{n!}}{\text{r}!(\text{n}-\text{r})!}$
${{^\text{n}}\text{C}_{\text{r}-1}}=\frac{(\text{n}-1)!}{(\text{r}-1)!(\text{n}-\text{1})-(\text{r}-1)!}$
$\frac{{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{\text{n!}(\text{r}-1)!(\text{n}-\text{r})!}{\text{r!}(\text{n}-\text{r})!(\text{n}-\text{1})!}$
$=\frac{\text{n}\times(\text{n}-1)!(\text{r}-1)!\times(\text{n}-\text{r}!)}{\text{r}\times(\text{n}-1)!(\text{r}-1)!(\text{n}-\text{r})!}$
$=\frac{\text{n}}{\text{r}}$
Hence proved
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Question 125 Marks
If ${^\text{16}}\text{C}_{\text{r}}={^\text{16}}\text{C}_{\text{r+2}},$ find ${^\text{7}}\text{C}_{4}.$
Answer
We have,
If ${^\text{n}}\text{C}_{\text{r}}={^\text{n}}\text{C}_{\text{p}}$
then r + p = n
16 = r + r + 2
r = 7
then ${^\text{n}}\text{C}_{\text{4}}={^\text{7}}\text{C}_{\text{4}}$
$\Rightarrow \frac{7!}{4!(7-4)!}$
$\Rightarrow \frac{7\times5\times6}{3\times2}$
$\Rightarrow 35$
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Question 135 Marks
If ${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}$ are in A.P. then find n.
Answer
${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}$ are in A.P.
$\therefore​​​​{^\text{n}}\text{C}_{\text{5}}-{^\text{n}}\text{C}_{\text{4}}={^\text{n}}\text{C}_{\text{6}}-{^\text{n}}\text{C}_{\text{5}}$
$\Rightarrow \frac{\text{n}!}{5!(\text{n}-5)!}-\frac{\text{n}!}{4(\text{n}-4)!}=\frac{\text{n}!}{6!(\text{n}-6)!}-\frac{\text{n}!}{6(\text{n}-5)!}$
$\Rightarrow \frac{\text{n}!}{5!(\text{n}-5)!}\Big[\frac{1}{5}-\frac{1}{\text{n}-4}\Big]=\frac{\text{n}!}{5!(\text{n}-6!)}\Big[\frac{1}{6}-\frac{1}{\text{n}-5}\Big]$
$\Rightarrow \frac{1}{\text{n}-5}\Big[\frac{\text{n}-4-5}{5(\text{n}-4)}\Big]=\frac{1}{5}\Big[\frac{\text{n}-5-6}{6(\text{n}-5)}\Big]$
$\Rightarrow \frac{\text{n}-9}{\text{n}-4}=\frac{\text{n}-11}{6}$
$\Rightarrow 6\text{n}-54=\text{n}^{2}-15\text{n}+44$
$\Rightarrow \text{n}^{2}-21\text{n}+98=0$
$\text{n}=7,14$
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Question 145 Marks
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Answer
There are total 9 courses are available and out of these 2 subjects are compulsory. So, Number of ways to select 2 compulsory and 3 option of 9 - 2 = 7 subject.
$={^{2}{\text{C}}}_{\text{2}}\times{^{7}{\text{C}}}_{\text{3}}$
$=1\times\frac{7\times6\times5}{3\times2}$
$=35$
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Question 155 Marks
If ${^\text{n+2}}\text{C}_{\text{8}}:{^\text{n-2}}\text{C}_{\text{4}},=57:16,$ Find n.
Answer
We have,
$\Rightarrow \frac{\frac{\text{n+2}!}{8!(\text{n-6})!}}{\frac{(\text{n-2})!}{(\text{n-6})!}}=\frac{57}{16}$
$\Rightarrow \frac{(\text{n+2})(\text{n+1})(\text{n})(\text{n-1})(\text{n}-2)}{8!(\text{n}-2)!}=\frac{57}{16}$
Cancelling (n - 2) from number and denominator
$\Rightarrow (\text{n}+2)(\text{n}+1)(\text{n})(\text{n}-1)=\frac{57\times7\times6\times5\times4\times3\times1\times16}{16}$
$\Rightarrow (\text{n+2})(\text{n}+1)(\text{n})(\text{n}-1)=21\times20\times19\times18$
Comparing both sides,
$\text{n}=19$
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Question 165 Marks
In how many ways can a football team of $11$ players be selected from $16$ players? How many of these will.
  1. Include $2$ particular players?
  2. Exclude $2$ particular players?
Answer
We have,
No of players $= 16$
No of players to be selected $= 11$
No of combination $={^\text{16}}\text{C}_{11}$
$=\frac{16!}{11!5!}$
$=\frac{16\times15\times14\times13\times12}{5\times4\times3\times2}$
$=4368$
  1. Include $2$ particular players:
Now we have to select $9$ more out of remaining $14$
Required number of ways $={^\text{14}}\text{C}_{9}$
$=\frac{14!}{9!5!}$
$=\frac{14\times13\times12\times11\times10}{5\times4\times3\times2}$
$=2002$
  1. Exclude $2$ particular players:
Now we have to select $11$ players more out of remaining $14$ players $={^\text{14}}\text{C}_{11}$
$=\frac{14!}{11!3!}$
$=\frac{14\times13\times12}{3\times2}$
$=364$
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Question 175 Marks
Evaluate $^{20}\text{C}_{5}+\sum\limits_\text{r=2}^5\ ^{25-\text{x}}\text{C}_4.$
Answer
We have,
$\Rightarrow\ ^{20}\text{C}_{5}+\sum\limits_\text{r=2}^5\ ^{25-\text{x}}\text{C}_4$
$\Rightarrow\ \big(^{20}\text{C}_{5}+{^{20}\text{C}_{4}}\big)+{^{21}\text{C}_{4}+^{22}}\text{C}_{4}+{^{23}\text{C}_{4}}$
$\Rightarrow\ \big(^{21}\text{C}_{5}+{^{21}\text{C}_{4}}\big)+{^{22}\text{C}_{4}}+{^{23}\text{C}_{4}}$
$\Rightarrow{^\text{23}}\text{C}_{\text{5}}+{^\text{23}}\text{C}_{\text{4}}$
$\Rightarrow{^\text{24}}\text{C}_{\text{5}}$
$\Rightarrow 42504$
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Question 185 Marks
In how many ways can one select a circket team of eleven from17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers?
Answer
We have,
There are total 5 bowlwrs and 12 batsm are avilable to select from.
Number of ways to select a team and 11 that includes exactly 4 bowlers.
= (7 batsm out of 12 bastsman) and (4 bowlers out of 5 bowlers)
$={^{12}{\text{C}}}_{\text{7}}\times{^{5}{\text{C}}}_{\text{4}}$
$\Rightarrow \frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}\times5$
$\Rightarrow 3960$
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Question 195 Marks
Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king?
Answer
We have,
Out of the 52 cards 4 are kings and 48 are Non-kings.
Five cards with at least one king
= (One king and 4 Non-kings) Or (two kings and 3 Non-kings) Or (3 kings and 2 Non-kings) Or (4 kings and 1 Non-kings)
$=\big({^{4}{\text{C}}}_{\text{1}}\times{^{48}{\text{C}}}_{\text{4}}\big)+\big({^{4}{\text{C}}}_{\text{2}}\times{^{48}{\text{C}}}_{\text{3}}\big)+\big({^{4}{\text{C}}}_{\text{3}}\times{^{48}{\text{C}}}_{\text{2}}\big)+\big({^{4}{\text{C}}}_{\text{4}}\times{^{48}{\text{C}}}_{\text{1}}\big)$
$=4\times\frac{48\times47\times46\times45}{4\times3\times2}+\frac{4\times3}{2}\times\frac{48\times47\times46}{3\times2}+4\times\frac{48\times47}{2}+1\times48$
$=778320+103776+4512+48$
$=886656$
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Question 205 Marks
A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Answer
We have, The number of ways selecting of 3 people out of 5 $\Rightarrow{^{5}{\text{C}}}_{\text{3}}=\frac{5!}{3!2!}$ $=\frac{5\times4}{2}=10$1 man can be selected from 2 men in ways and 2 wimen can be selected fron 3 ways,
The required number of committees
$={^{2}{\text{C}}}_{\text{1}}\times{^{3}{\text{C}}}_{\text{2}}$ $=\frac{2!}{1!1!}\times\frac{3!}{2!1!}$ $=6$
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Question 215 Marks
Out of $18$ points in a plane, no three are in the same straight line except five points which are collinear. How many:
  1. Straight lines.
  2. Triangles can be formed by joining them?
Answer
There are $8$ points in a plane out of which $5$ points are collinear.
Then number of strianght lines joining these points are,
$\Rightarrow {^\text{n}}\text{C}_{\text{2}}-({^\text{p}}\text{C}_{\text{2}}-1)$
$\Rightarrow {^\text{n}}\text{C}_{\text{2}}-{^\text{p}}\text{C}_{\text{2}}+1$
$\Rightarrow{^\text{18}}\text{C}_{\text{2}}-{^\text{5}}\text{C}_{\text{2}}+1$
$\Rightarrow \frac{18\times17}{2}-\frac{5\times4}{2}+1$
$\Rightarrow144$
Number of triangle $={^\text{13}}\text{C}_{\text{3}}$
$=\frac{13!}{3!10!}=\frac{13\times12\times11}{3\times2}$
$=13\times2\times11$
$=13\times22$
$=806$
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Question 225 Marks
Let r and n be positive integers such that 1 < r < n. Then prove the following:
${^\text{n}\text{C}}_{\text{r}}+2\ {^\text{n}\text{C}}_{\text{r}-1}+{^\text{n}\text{C}}_{\text{r}-2}={^\text{n+2}\text{C}}_{\text{r}}$
Answer
We have,
$\text{L.H.S.}={^\text{n}\text{C}}_{\text{r}}+2\ {^\text{n}\text{C}}_{\text{r}-1}+{^\text{n}\text{C}}_{\text{r}-2}$
$=({^\text{n}\text{C}}_{\text{r}}+{^\text{n}\text{C}}_{\text{r}-1})+({^\text{n}\text{C}}_{\text{r}-2}+{^\text{n}\text{C}}_{\text{r}-1})$
$={^\text{n+1}\text{C}}_{\text{r}}+{^\text{n+1}\text{C}}_{\text{r}-1}$
$=(\text{n}+1)+{^\text{1}\text{C}}_{\text{r}}$
$={^\text{n+2}\text{C}}_{\text{r}}$
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Question 235 Marks
In an examination, a student has to answer 4 questions out of 5 questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Answer
Total number of quation = 5
Total number of quatin to be answered = 4
Given thet 1 and 2 quation are compulsory, the number of ways in which a student can choose the quations will follow the following way.
Total quation = 5 - 2 = 3
Out of 3 remaining questions a student has to select any 2 for answer.
$\Rightarrow {^{3}{\text{C}}}_{\text{2}}=3$
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Question 245 Marks
Find the number of permutations of n different things taken r at a time such that two specified things occur together?
Answer
There are x things
Two specific things are to occur together, so remaining things are (r - 2).
Now, number of ways to arrange (r - 2) things out of $(\text{n}-2)={^\text{n-2}}\text{C}_{\text{r-2}}$
Two things can be arranged is (r - 1) ways.
These two can be placed in 2 ways.
Therefore, Required number of ways $2(\text{r}-1)=2(\text{r-1}){^\text{n-2}}\text{C}_{\text{r-2}}$
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Question 255 Marks
Let r and n be positive integers such that 1 < r < n. Then prove the following:
$\frac{{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{\text{n}-\text{r}+1}{\text{r}}$
Answer
We have,
${{^\text{n}}\text{C}_{\text{r}}}=\frac{\text{n!}}{\text{r}!(\text{n}-\text{r})!}$
${{^\text{n}}\text{C}_{\text{r}-1}}=\frac{\text{n}!}{(\text{r}-1)!(\text{n}-\text{r}+1)!}$
$\frac{{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{\text{n!}(\text{r}-1)!(\text{n}-\text{r}+1)!}{\text{r!}(\text{n}-\text{r})!(\text{n}-\text{r})!}$
$=\frac{\text{}(\text{r}-1)!(\text{n}-\text{r}+1)!\times(\text{n}-\text{r})!}{\text{r}_{2}\times(\text{r}-\text{1})!(\text{n}-\text{r})!}$
$=\frac{\text{n}-\text{r}+1}{\text{r}}$
Hence proved
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Question 265 Marks
A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes in how many ways can the teams be constituted?
Answer
Total number of student is XI = 20
Total number of student is XII = 20
Total number of student to be selected is C team = 11
${^{20}{\text{C}}}_{\text{5}}\times{^{20}{\text{C}}}_{\text{6}}+{^{20}{\text{C}}}_{\text{6}}\times{^{20}{\text{C}}}_{\text{5}}$
$=2\Big({^{20}{\text{C}}}_{\text{6}}\times{^{20}{\text{C}}}_{\text{5}}\Big)$
$=2\Big(\frac{20!}{6!14!}\times\frac{20!}{5!15!}\Big)$
$=\frac{2\times20\times19\times18\times17\times16\times15\times20\times19\times18\times17\times16}{6\times5\times4\times3\times2\times5\times4\times3\times2\times1}$
$=19\times17\times16\times15\times2\times19\times3\times17\times8$
$=1201870080$
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Question 275 Marks
How many triangles can be obtained by joining 12 points, five of which are collinear?
Answer
We have,
Since 5 out of 12 points are collinear, So the number of triangle will be,
$={^{12}{\text{C}}}_{\text{3}}-{^{5}{\text{C}}}_{\text{3}}$
$=\frac{12!}{3!9!}-\frac{5!}{3!2!}$
$=\frac{12\times11\times10}{3\times2}-\frac{5\times4}{2}$
$=220-10$
$=210$
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Question 285 Marks
Find the number of ways of selection 9 balls from 6 red balls, 5 white balls and 5 balls if each selection consists of 3 balls of each colour.
Answer
There are 6 red balls, 5 white balls and 5 blue balls.
Number of ways to select 9 balls consisting of 3 balls of each colour.
= (3 red out of 6 red) and (3 white out of 5 white) and (3 blue out of 6 blue balls)
$={^{6}{\text{C}}}_{\text{3}}\times{^{5}{\text{C}}}_{\text{3}}\times{^{5}{\text{C}}}_{\text{3}}$
$=\frac{6\times5\times4}{3\times2\times1}\times\frac{5\times4}{2}\times\frac{5\times4}{2}$
$=2000$
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Question 295 Marks
There are $10$ professors and $20$ students out of whom a committee of $2$ professors and $3$ students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees:
  1. A particular professor is included.
  2. A particular student is included.
  3. A particular student is excluded.
Answer
We have,
Total number of professor $= 10$
Total number of student $= 20$
Committee of $2$ professor and $3$ student can be selected in $={^\text{10}}\text{C}_{2}\times{^\text{20}}\text{C}_{3}.$
$=\frac{10!}{2!8!}\times\frac{20!}{3!17!}$
$=\frac{10\times9}{2}\times\frac{20\times19\times18}{3\times2}$
$=51300$
  1. A particular professor is included:
Committee is ${^\text{9}}\text{C}_{1}\times{^\text{20}}\text{C}_{3}$
$=\frac{9!}{8!}$
$=\frac{20}{3!\times17!}$
$=\frac{9\times20\times19\times18}{3\times2}$
$=10260$
  1. A particular student is included:
Committee is ${^\text{10}}\text{C}_{2}\times{^\text{19}}\text{C}_{2}$
$=\frac{10!}{2!8!}\times\frac{19!}{2!17!}$
$=\frac{10\times9\times19\times18}{2\times2\times1}$
$=7695$
  1. A particular student is excluded:
Committee is ${^\text{10}}\text{C}_{2}\times{^\text{19}}\text{C}_{3}$
$=\frac{10!}{2\times8!}\times\frac{19!}{3!16!}$
$=\frac{10\times9\times19\times18\times17}{2\times3\times2}$
$=43605$
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Question 305 Marks
Let r and n be positive integers such that 1 < r < n. Then prove the following:
$\text{n}\ {{^\text{n-1}}\text{C}_{\text{r-1}}}=(\text{n}-\text{r}+1){{^\text{n}}\text{C}_{\text{r}-1}}$
Answer
We have,$\text{n}\times {{^\text{n-1}}\text{C}_{\text{r-1}}}$
$=\text{n}\times\frac{(\text{n}-1)!}{(\text{r}-1)!(\text{n}-\text{r})!(\text{r}-1)!}$
$=\frac{{\text{n}!\times(\text{n}-\text{r}+1)}}{(\text{r}-1)!(\text{n}-\text{r})!(\text{n}-\text{r}+1)}$
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
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Question 315 Marks
If ${^8}\text{C}_{\text{r}}-{^7}\text{C}_{\text{3}}={^7}\text{C}_{\text{2}},$ Find r.
Answer
Applying formula ${^\text{n}\text{C}_{\text{r}}}=\frac{\text{n}!}{\text{r!}(\text{n}-\text{r})!}$
$\frac{8!}{\text{r!}(8-\text{r})!}=\frac{7!}{2!5!}+\frac{7!}{3!4!}$
$\frac{8\times7!}{\text{r}!(8-\text{r})!}=\frac{7!}{2\times5\times4!}+\frac{7!}{3\times2\times4!}$
$\frac{8\times7!}{\text{r}!(8-\text{r})!}=\frac{7!}{2\times5\times4!}\Big(\frac{1}{5}+\frac{1}{5}\Big)$
Cancelling from both
$\frac{8\times7!}{\text{r}!(8-\text{r})!}=\frac{8}{2\times15\times4!}$
Cancelling 8 from both sides
2 × 5 × 3 × 4 × 3 × 2 × 1 = r! (8 - r)!
⇒ r = 3
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Question 325 Marks
A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular side and two on the other side. In how many ways can they be seated?
Answer
No of persons = 16
Conditionon persons = 4 and 2 = 6
Remaining people = 16 - 6 = 10
So, lets fill 8 people on both sides first sides from these 10.
First side, we can select 4 out of 10.
$={^\text{10}}\text{C}_{\text{4}}\times{^\text{6}}\text{C}_{\text{6}}$
Now, we can arrange these 8 people on both sides in 8! × 8! ways.
Answer $={^\text{10}}\text{C}_{\text{4}}\times{^\text{6}}\text{C}_{\text{6}}(\times8!)^{2}$
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Question 335 Marks
Determine the number of 5 card combination out of a deck of if there is one ace in each combination.
Answer
We have,
Out of 52 cards 4 are ace and 48 are Non-ace.
Number of ways to select 5 cards with exacly one ace.
= (1 Non-ace out of 4 ace) and (4 Non-ace out of 48 Non-ace)
$={^{4}{\text{C}}}_{\text{1}}\times{^{48}{\text{C}}}_{\text{4}}$
$=4\times\frac{48\times47\times46\times45}{4\times3\times2\times1}$
$=778320$
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Question 345 Marks
A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has:
  1. No girl?
  2. At least one boy and one girl?
  3. At least $3$ girls?
Answer
  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.
$\Rightarrow {^{7}{\text{C}}}_{\text{5}}=\frac{7!}{5!2!}$
$=\frac{6\times7}{2}=21$
  1. 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,
$={^{7}{\text{C}}}_{\text{1}}\times{^{4}{\text{C}}}_{\text{4}}+{^{7}{\text{C}}}_{\text{2}}\times{^{4}{\text{C}}}_{\text{3}}+{^{7}{\text{C}}}_{\text{3}}\times{^{4}{\text{C}}}_{\text{2}}+{^{7}{\text{C}}}_{\text{4}}\times{^{4}{\text{C}}}_{\text{1}}$
$=7+84+210+140$
$=441$
  1. 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,
$={^{4}{\text{C}}}_{\text{3}}\times{^{7}{\text{C}}}_{\text{2}}+{^{4}{\text{C}}}_{\text{4}}\times{^{7}{\text{C}}}_{\text{1}}$
$=84+7$
$=91$
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Question 355 Marks
Find the number of ways in which:
  1. A selection.
  2. An arrangement, of four letters can be made from the letters of the word $\text{'PROPORTION'.}$
Answer
The given word is $\text{'PROPORTION'.}$
Total letters $= 10$
Number of $P = 2,$ Number of $R = 2$
Number of $O = 3,$ Number of $T = 1$
Number of $I = 1,$ Number of $N = 1$
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
$= 15 + 30 + 3 + 5$
$= 53$
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 $= 758$
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Question 365 Marks
A parallelogram is cut by two sets of m lines parallel to its sides. Find the number of parallelograms thus formed.
Answer
In a parallelogram, there are 2 sets of parallel line. Each set of parallel lines consists of lines and each parallelogram is formed by choosing two line from the first and two straing lines the second set.
Hence, the total nmber of parallelogram $={^\text{m+2}}\text{C}_{\text{2}}\times{^\text{m+2}}\text{C}_{\text{2}}$
$=({^\text{m+2}}\text{C}_{\text{2}}){^{2}}$
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Question 375 Marks
How many words can be formed by taking $4$ letters at a time from the letters of the word $\text{'MORADABAD' ?}$
Answer
The given word is $\text{'MORADABAD'.}$
Number of $M = 1,$ Number of $O = 1$
Number of $R = 1,$ Number of $A = 3$
Number of $D = 2,$ Number of $B = 1$
  1. Number of arrangement of $4$ letters.
Selected from these $={^\text{6}}\text{C}_{\text{4}}\times4!$
$=15\times24$
$=360$
  1. 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
$={^\text{2}}\text{C}_{\text{1}}\times​​{^\text{5}}\text{C}_{\text{2}}\times\frac{4!}{2!}$
$=2\times10\times12$
$=240$
  1. Two ailke and with more than
Number of ways to arrange therefour
$={^\text{2}}\text{C}_{\text{2}}\times​​{^\text{5}}\text{C}_{\text{2}}\times\frac{4!}{2!2!}$
$=6$
  1. 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 $= 626$
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Question 385 Marks
A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions?
Answer
Total number of quation = 12
Total number of quatin to be answered = 7
Each group has 6 quation more than 5 quation from either grou is not permitted, the number of ways a student can choose quation,
${^{6}{\text{C}}}_{\text{2}}\times{^{6}{\text{C}}}_{\text{5}}+{^{6}{\text{C}}}_{\text{4}}\times{^{6}{\text{C}}}_{\text{4}}+{^{6}{\text{C}}}_{\text{4}}\times{^{6}{\text{C}}}_{\text{3}}+{^{6}{\text{C}}}_{\text{5}}\times{^{6}{\text{C}}}_{\text{2}}$
$=2\Big({^{6}{\text{C}}}_{\text{2}}\times{^{6}{\text{C}}}_{\text{5}}+{^{6}{\text{C}}}_{\text{3}}\times{^{6}{\text{C}}}_{\text{4}}\Big)$
$=2\Big(\frac{6!}{2!4!}\times\frac{6!}{5!1!}+\frac{6!}{3!3!}\times\frac{6!}{4!2!}\Big)$
$=\frac{2\times6\times5\times6}{2}\Big(1+\frac{20}{6}\Big)$
$=30\times26=780$
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Question 395 Marks
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Answer
We have,
Bag contains 5 black and 6 red balls.
Number of ways to select 2 black balls out of 5 black and 3 red balls out of 6 red balls.
$={^\text{5}}\text{C}_{\text{2}}\times{^\text{6}}\text{C}_{\text{3}}$
$=\frac{5\times4}{2}\times\frac{6\times5\times4}{3\times2}$
$=200$
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Question 405 Marks
Find the number of:
  1. Diagonals.
  2. Triangles formed in a decagon.
Answer
We have, A decagon has $10$ sides By joining any two angular points We get a line which is either a side or a diagonal Number of line,
$\Rightarrow{^{10}{\text{C}}}_{\text{2}}=\frac{10!}{2!8!}$
$=\frac{10\times9}{2}$
$=45$
Number of sides $= 10$
Number of diagonals $= 45 - 10 = 35$
Also, by joining $3$ angular points a triangle in formed
$={^{10}{\text{C}}}_{\text{3}}$
$=\frac{10!}{3!7!}$
$=\frac{10\times9\times8}{3\times2}$
$=\frac{720}{6}$
$=102$
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