Exercise 3.6
1.Find the value of
$a)^{15}C_4 $
Answer-Here n=15, r=4
as,$^nC_r=\frac{n}{r!(n-r)!}$
$=\frac{15!}{4!(15-4)!}$
$=\frac{15.14.13.12.11!}{4.3.2.1}$
$=5.7.13.3$
=1365
b)$^{80}C_2$
Answer- n=80 ,n=2
as, $^nC_r=\frac{n}{r!(n-r)!}$
=$\frac{80!}{2!(80-2)!}$
$=\frac{80!}{2!78!}$
$=\frac{80\times79\times78!}{2\times78!}$
=40.79
=3260
$c)^{15}C_4+^{15}C_4$
Answer-
As, $^nC_r+^nC_r=^{n+1}C_r$
So, n=15, r=4
$\therefore ^{15}C_4+^{15}C_4$
$=^{16}C_5$
d)$^{20}C_{16}+^{19}C_{16}$
Answer-
$=^{19}_C16+^{19}C_15-{19}C_16$
$=^{19!}C_{15}$
$=\frac{19!}{(19-15)!}$
$=\frac{19!}{15!4!}$
$=\frac{19.18.17.16.15!}{15!.4.3.2.1}$
=19.6.17.2
=3876
3.Find r if
a)$^{14}C_{2r}:^{10}C_{2r-4}=\frac{143}{10}$
Answer-
$\frac{\frac{14!}{(2r)!(4-2r)!}}{\frac{10!}{((2r-4)!(14-2r)!}}$=$\frac{143}{10}$
$\frac{14!}{(2r)!(4-2r)}!$$\times$$\frac{((2r-4)!(14-2r)!}{10!}$
$\frac{14.13.12.11.10!(2r-4)!}{2r(2r-1)(2r-2)(2r-3)(2r-4)!10!}=143:10$
$\frac{14.13.12.11}{2r(2r-1)(2r-2)(2r-3)}=143:10$
$r(2r-1)(2r-2)(2r-3)=\frac{14.13.12.11.10}{143.2}$
$r(2r-1)(2r-2)(2r-3)=720$
$r(2r-1)(2r-2)(2r-3)=10.9.8$
$r=10$
4. Find n and r if
a) $^{n}P_{r}$ and $^{n}C_{n-r}=120$
Answer-
$^nP_r$ $^nC_n-r=120$
$\frac{n!}{(n-r)!}=720 ....(i) \frac{n!}{(n-r)!r!}=120$ ....(ii)
dividing by (i) and (ii)
$\frac{n!}{(n-r)!}\times\frac{r!(n-r)!}{n!}=\frac{720}{120}$
r!=6
so r=3
b)$^nC_{r-1}:^nC_r:^nC_{r+1}=20:35:42$
Answer-
As, $^nC_{r-1}:^nC_r=\frac{20}{35}$
$\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}$=$\frac{20}{35}$
$\frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n-r)!}{n!}$=$\frac{20}{35}$
$\frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)!(n-r)!}$=$\frac{20}{35}$
$\frac{r}{n-r+1}=\frac{20}{35}$
35r=20(n-r+1)
11r-4n=4 .....(i)
now, $^nC_r:^nC_{r+1}=35:42$
$\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-(r+1))!}}$=$\frac{35}{42}$
$\frac{n!}{r!(n-r)!} \times \frac{(r+1)!(n-(r+1))!}{n!}$=$\frac{35}{42}$
$\frac{(r+1)r!(n-r-1)!}{r!(n-1)(n-r-1)!}$=$\frac{35}{42}$
$\frac{r+1}{n-r}=\frac{35}{42}$
$\therefore$ 42(r+1)=35(n-r)
$\therefore$ 42r+42=35n-35r
11r-5n=-6 ....(ii)
So, 11r-4n=4
11r-5n=-6
solving above equation we get the value of n=10
and now substituting the value of n= 10 in eq (i) we get,
11r-5(10)=-6
11r=44
r=4 and n=10
7.Find the number of ways of drawing 9 balls from a bag that has 6 red balls 8 green balls and 7 blue was so that three balls of every colour are drawn.
Answer-
Total number ways of balls=8+5+7=18
Number of red balls=6
Number of green balls=5
Number of blue balls=7
9 balls are to be drawn 3 of each colour,
No. of ways of drawing 3 red balls of 6 red balls=$^6C_3$
similarly 3 green balls out of 5 geen balls=$^5C_3$
3 blue ball out of 7 blue balls=$^7C_3$
\therefore $total number ways$=$^6C_3\times^5C_3\times^7C_3$
=$\frac{6!}{3!3!}\times \frac{5!}{3!2!} \times \frac{7!}{3!4!}$
$=20 \times 10 \times 35 $
=7000
8.It. find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Answer-
there are 6 boys and 4 girls
we have to select 3 boys out of 6 boys=$^6C_3$
we have to select 2 girls out of 4 girls=$^4C_2$
$\therefore$ total number of ways
=$^6C_3 \times ^4C_2$
$=\frac{6!}{3!3!} \times \frac{4!}{2!2!}$
$=20\times 6$
9. After meeting every participant shake hands with every other participant. If the number of handshakes is 66 find the number of participants in the meeting.
Answer-
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = nC2.
Given 66 handshakes were exchanged.
∴ 66 = nC2
∴ 66 =$\frac{n!}{2!(n-2)!}$
∴ 66 × 2 =$\frac{n(n-1)(n-2)!}{(n-2)!}$
∴ 132 = n(n – 1)
∴ n(n – 1) = 12 × 11
Comparing on both sides, we get
n = 12
∴ 12 participants were present at the meeting.
10. If 20 points are marked on a circle how many chords can be drawn?
Answer-
There are 20 points on a circle.To draw a chord, 2 points are required.
∴ the number of chords that can be drawn through 20 points on the circle.
= 20C2
$=\frac{20!}{{2!18!}$
$=\frac{20×19×18!}{{2×1×18!}$
= 190.
10. If 20 points are marked on a circle how many chords can be drawn?
Answer-
There are 20 points on a circle.To draw a chord, 2 points are required.
∴ the number of chords that can be drawn through 20 points on the circle.
= 20C2
$=\frac{20!}{{2!18!}$
$=\frac{20×19×18!}{{2×1×18!}$
= 190.
11.find the number of diagonals of an n sided polygon. In particular, find the number of diagonals when.
a) n=10 b)n=15 c)n=12 d)n =8
Answer-
a)In an n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw nC2 lines including sides.
∴ Number of diagonals in n sided polygon.
= nC2 – n .........(n = number of sides)
n = 10
nC2 – n = 10C2 – 10
$\frac{10.9}{1.2}-10$
= 45 – 10
= 35
b)There are n vertices in the polygon of n-sides.
If we join any two vertices, we get either side or the diagonal of the polygon.
Two vertices can be joined in nC2 ways.
∴ total number of sides and diagonals = nC2 But there are n sides in the polygon.
∴ total number of the diagonals = nC2 – n
n = 15 sides
∴ the number of diagonal that can be drawn.
= 15C2 – 15
=$\frac{15×14×13!}{2×13!}-15$
=15×142-15
= 105 – 15
= 90
c)In an n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw nC2 lines including sides.
∴ Number of diagonals in n sided polygon.
= nC2 – n ...(n = number of sides)
n = 12,
nC2 – n = 12C2 – 121
$=\frac{2×11}{1×2}-12$
= 66 – 12
= 54
d)There are n vertices in the polygon of n-sides.
If we join any two vertices, we get either side or the diagonal of the polygon.
Two vertices can be joined in nC2 ways.
∴ total number of sides and diagonals = nC2
But there are n sides in the polygon.
∴ total number of the diagonals = nC2 – n
n = 8 sides
∴ the number of diagonals that can be drawn
12.There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Answer-
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
∴ they intersect at a point
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent.
= 20C2
$=\frac{20!}{2!18!}$
$=\frac{20×19×18!}{2×1×18!}$
= 190
13.10 points are plotted on a plane. Find the number of straight lines obtained by joining these points if
A)no three points are collinear.
B) four points are collinear.
Answer-
A )line is drawn by joining 2 points from the given 10 points.
∴ number of straight lines= 10C2
=$\frac{10!}{2!8!}$
$=\frac{10×9×8!}{2×8!}$
= 45, if no three points are collinear.
B) when four points are collinear then the total number of lines= 10C2 but it contains lines using that 4 collinear points make only one line
lines using 4 points = 4C2
No. of ways is 4 points are collinear = 10C2 - 4C2 +1
= 45 - $\frac{4!}{2!2!}$+1
=46- 6
=40
14.Find the number of triangles formed by joining 12 points if a) no three points are collinear
b) 4 points are collinear.
Answer-
a)A triangle can be formed by selecting 3 non-collinear points
3 points be selected from 12 points
$^{12}C_3=\frac{12!}{9!3!}$
$=\frac{12.11.10.9!}{9!3!}$
=220 ways
b)There are 12 points on the plane
when 4 of these points are collinear
$ \therefore $ number of triangles that can be from these points=$^{12}C_3-^{4}C_3$
=220-$\frac{4!}{3!1!}$
=220-4
=216
15. A word has three consonant 8 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are use and?
Answer-
4 consonants can be selected from 8 consonants in $^8C_4$ ways 2 vowels can be selected from 3 vowels in $^3C_2$
$\therefore$ number of words with 4 consonants and 2 vowels
=$^8C_4\times ^3C_2 $
=$\frac{8!}{4!4!}\times \frac{3!}{2!1!}$
=70 $\times$ 3
=210
now, each of these 6 letters can be arranged in 6P6 ways=6!
$\therefore$ total numbers can be formed 4 consonants and 2 vowels
$=210 \times 6!$
=151200
16. Find n if,
(i) ^nC_8=^nC_{12}
Answer-
$^nC_{12}=^nC_8$ $^nC_x=^nC_y$ then either x=y or x=n-y
\therefore 12=n-8 \hspace{2cm} \because 12\neq 8
n=20
(ii)$^{23}C_{3n}=^{23}C_{2n+1}$
Answer-
$^{23}C_{3n}=^{23}C_{2n+3}$ \because $^nC_x=^nC_y$ then either x=y or x=n-y
\therefore 3n=2n+3 or 3n+2n+3=23
$n=3$ or $5n=20$
\therefore $n=3$ or $n=4$
(iii)$^{21}C_{6n}=^{21}C_{n^2-5}$
Answer-
$^{21}C_{6n}=^{21}C_{n^2-5}$
$\because$ $^nC_x=^nC_y$ then either $x=y or x=n-y$
$\therefore$ $6n=n^2+5$ or $6n=21-(n^2+5)$
$\therefore$ $n^2-6n+5=0$ or $6n=21-n^2-5$
$\therefore$ $(n-1)(n-5)=0$ or $n^2+6n-16=0$
$\therefore$ n=1 or n=5 or n=-8 or n=2
if n=5 then $n^2+5=30>21$ or $ n\neq-8 \\ so, n=2$\\
(iv)^{2n}C_{r-1}=^{2n}C_{r+1}
Answer-
$^{2n}C_{r-1}=^{2n}C_{r+1}$
$\therefore$ $ ^nC_x=^nC_y$ then either x=y or x=n-y
\therefore r-1=r+1 or r-1=2n-r-1
\therefore r=1+1 or r-1+r+1=2n
\therefore r=2 or r=n
(v) $^{n}C_{n-2}=15$
Answer-
$^{n}C_{n-2}=15$
$\frac{n!}{(n-2)!(n-n+2)!}$=15
$\frac{n(n-1)(n-2)!}{(n-2)!2!}$=15
$\frac{n(n-1)}{2}=15$
n(n-1)=15.2
n(n-1)=30
n(n-1)=6.5
so, n=6
17. Find x if $^{n}P_{r}=x$ \times ^nC_r
Answer-
$^{n}P_{r}=x$ \times $^nC_r$
$\frac{n!}{(n-r)!}=x$ \times $\frac{n!}{r!(n-r)!}$
$1=x \times \frac{1}{r!}$
x=r!
18. Find r if
$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7=$^{14}C_r$
Answer-
$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7=^{14}C_r$
As, $^nC_r+^nC_{r-1}=^{n+1}C_r$
$^{14}C_r$=$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7$
$=^{11+1}C_5+^{12}C_6+^{13}C_7$
$=^{12}C_5+^{12}C_6+^{13}C_7$
$=^{12+1}C_6+^{13}C_7$
$=^{13}C_6+^{13}C_7$
$=^{14}C_7$
so, r=7
19. Find the value of $\sum_{r=1}^{4}{} ^{21-r}C_4$\\
Answer-
$^{21-r}C_4$
$=^{21-1}C_4 + ^{21-2}C_4+ ^{21-3}C_4+^{21-4}C_4$
$=^{20}C_4+^{19}C_4+{18}C_4+{17}C_4$
$\because$ $^nC_r+^nC_{r-1}$=$^{n+1}C_r$
$^nC_{r-1}=^{n+1}C_r-^nC_r$
$=^{20}C_4+^{19}C_4+^{18}C_4+{18}C_5-{17}C_5$
$=^{20}C_4+\left(^{19}C_4+^{19}C_5\right)-{17}C_5$
$=^{20}C_4+^{19+1}C_5-{17}C_5$
$=^{20}C_4+^{20}C_5-{17}C_5$
$=^{21}C_5-{17}C_5$
$=\frac{21!}{5!(21-5)!}-\frac{17!}{5!(17-5)!}$
$=\frac{21!}{5!16!}$-$\frac{17!}{5!12!}$
$=\frac{21.20.19.18.17.16!}{5!16!}$-$\frac{17.16.15.14.13.12!}{5!12!}$
$=\frac{21.20.19.18.17}{5.4.3.2.1}$-$\frac{17.16.15.14.13}{5.4.3.2.1}$
= 20349 - 6188
=1416
20. Find the differences between the greatest values in the following:
(a) $^{14}C_r and ^{12}C_r
Answer-
here n=14 is an even
$^{14}C_r$ then the greatest value of nCr occurs at $r=\frac{n}{2}$
$r=\frac{14}{2}$
r=7
Greatest value of $^{14}C_r=^{14}C_7$
$=\frac{14!}{(14-7)!}$
$=\frac{14.13.12.11.10.9.8.7!}{7!(14-7)!}$
$=\frac{14.13.12.11.10.9.8}{7.6.5.4.3.2.1}$
=3432
Also, for the greatest values of ^{12}C_r $
here, n=12 $r=\frac{12}{2}=6$
$^{12}C_6$
$=\frac{12!}{6!(12-6)!}$
$=\frac{12.11.10.9.8.7.6!}{6!6!}$
$=\frac{12.11.10.9.8.7}{6.5.4.3.2.1}$
=924
the difference between
=$^{14}C_r$ and $^{12}C_r$
=3432-964
=2508
(b)$^{13}C_r$ and $^{8}C_r$
Answer-
the greatest value of
$^nC_r$ $r=\frac{n}{2}$ if n is even and if n is odd $r=\frac{n-1}{2}$
$^{13}C_r=$
here n=13 $r=\frac{13-1}{2}$
$r=\frac{12}{2}=6$
$^{13}C_r={13}C_6$
=$\frac{13!}{6!7!}$
=$\frac{13.12.11.10.9.8.7!}{7!6.5.4.3.2.1}$
=1716
so,$^8C_r$ here n=8
n=\frac{8}{2}=4
$^8C_4$
$=\frac{8!}{4!6!}$
$=\frac{8.7.6.5.4!}{4!6!}$
=70
$\therefore$ the difference between
=$^{13}C_r$ and $^{8}C_r$
=1716-70
=1646
(c)$^{15}C_r$ and $^{11}C_r$
Answer-
the greatest value of
$^nC_r$ $r=\frac{n}{2}$ if n is even and if n is odd $r=\frac{n-1}{2}$
$^{15}C_r$
here, n=15 so $r=\frac{n-1}{2}=\frac{15-1}{2}=7$
$^{15}C_r$=$^{15}C_7$
$=\frac{15!}{7!8!}$
$=\frac{15.14.13.12.11.10.9.8!}{8!.7.6.5.4.3.2.1}$
=6435
$^{11}C_r$
here, n=11 so $r=\frac{11-1}{2}=5$
$^{11}C_r$=$^{11}C_5$=$\frac{11!}{5!6!}$
$=\frac{11.10.9.8.7.6!}{6!5.4.3.2.1}$
=462
the difference between $^{15}C_r$ and $^{11}C_r$
=6435-462
=5973
21.In how many ways can a boy invite his five friends to a party so that at least 3 join the party?
Answer-
Number of friends n=5
Atleast 3 friends join party
$^5C_3+^5C_4+^5C_5$
$=\frac{5!}{3!(5-3)!}+\frac{5!}{4!}+1$
$=10+5+1$
=16
22. A group consists of 9 men and 6 women.A team of 6 is to be selected.How many possible selections will have atleast three women.
Answer-
There are 9 men and 6 women
will have 3 women atleast
=3W3M+4W2M+5W1M+6W no men
=$^6C_3 \times ^9C_3+^6C_4 \times ^9C_2+^6C_5 \times ^9C_1+1$
=1680+540+54+1
=2275
Therefore, 2275 can be formed if the team consist atleast 3 women
23. A committee of 10 person is to be formed from a group 10 women 8 men.How many possible committees will have at least five women? How many possible committee will have men in majority?
Answer-
There are 10 women and 8 men.
A committee of 10 persons is to be formed.If atleast 5 women have to be selected in a committee
so, possible selection are
5 Women 5 Men 6 Women 4 Men
7 Women 3 Men 8 Women 2 Men
9 Women 1 Men 10 Women 0 Men
$\therefore$ the number of ways of forming committees such that at least 5 women are included
$={10}C_5 \times 8C_5+{10}C_6 \times 8C_4+{10}C_7 \times 8C_3+{10}C_8 \times 8C_2+{10}C_9 \times 8C_1+{10}C_{10}\times 8C_0$
=252.56+210.70+120.56+45.28+80+1
=36873
Men in Majority
men in majority means 6,7,8
=$^8C_6 \times ^{10}C_4+^8_7 \times ^{10}C_3+^8C_8 \times ^{10}C_2$
=5880+960+45
=6885
24. A question paper has two sections, section 1 has 5 questions and section 2 has 6 questions a student must answer at least two questions from each section among six question hi answers. How many different choices does the student have in choosing questions?
Answer-
There are 11 question out of which section 1 has 5 question and section 2 has 6 section\\
possible choices S I-2 S II-4, S I-3 S II-3 S I-4 S II-2
$^5C_2 \times ^6C_4+^5_3 \times ^6C_3+^5C_4 \times ^6C_2$
=$10 \times 15+10 \times 20+5 \times 15$
=150+200+75
=425
$\therefore$ in 425 ways student can select 6 questions taking atleast 2 question from each
25.There are 5 wicket keepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicket keeper and atleast 4 bowlers in the team how many different teams can be formed.
Answer-
Among 22 players there are 3 wicketkeepers+5 bowlers=8players and the remaining 22-8=14 players batsmen.
i)1 wicketkeeper 4 bowlers 6 players
ii)1 wicketplayer 5 bowler 5 players
now, number of selection in
(i)$^3C_1 \times ^5C_4 \times {14}C_6$+$^3C_1 \times ^5C_5 \times {14}C_5$
$=3 \frac{5!}{4!} \frac{14!}{6!9!}+3 \frac{14!}{5!9!}$
=45045+6006
=51051
26.5 students are selected from 11 how many ways can the students be selected if\\
A) two specified students are selected?
B) to specified student are not selected?
Answer-
A)5 students are to be selected from 11 students
when 2 specified students are always together
then remaining student can be selected from 11-2=9 students
$\therefore$ number of ways of selecting 3 students from 9 students
= $^9C_3$
=$\frac{9!}{3!6!}$
=84
$\therefore$ selection is done in 84 ways when 2 specified students are included
B)Number of students is 11 we have to select 5 students
2 particular student are not selected
if two particular student are not selected then from remaining (11-2)=9 students, we have to select 5 students
This can be done$ $^9C_5$ ways
$\therefore$ total number of selections
$^9C_5$
$=\frac{9!}{5!4!}$
$=\frac{9.8.7.6.5!}{5!4!}$
=196
Thanks for reading inlogical math
Permutation Combination Exercise 3.1
Permutation Combination Exercise 3.2
Permutation Combination Exercise 3.3
Permutation Combination Exercise 3.4
Permutation Combination Exercise 3.5
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| Permutation and Combination Exercise 3.6 11th Class MH Board |
Permutation and Combination Exercise 3.6 11th Class MH Board
