A committee of 7 has to be formed from 9 boys and 4 girls.
Question: 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 (i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls? Solution: A committee of 7 has to be formed from 9 boys and 4 girls. I. Exactly 3 girls: If there are exactly 3 girls in the committee, then there must be 4 boys, and the ways in which they can be chosen is $={ }^{4} \mathrm{C}_{3}$ $x$ ${ }^{9} \mathrm{C}_{4}$ II. At least 3 girls: Here the possib...
Read More →Out of 6 teachers and 8 students, a committee of 11 is being formed.
Question: Out of 6 teachers and 8 students, a committee of 11 is being formed. In how many ways can this be done, if the committee contains (i) exactly 4 teachers? (ii) at least 4 teachers? Solution: Since the committee of 11 is to be formed from 6 teachers and 8 students. (i) Forming a committee with exactly 4 teachers Choosing 4 teachers out of 6 in ${ }^{6} \mathrm{C}_{4}$ ways. Remaining 7 from 8 students in ${ }^{8} \mathrm{C} 7$ ways. Thus, total ways in (i) are ${ }^{6} \mathrm{C}_{4} \ti...
Read More →In an examination, a candidate is required to answer 7 questions out of 12,
Question: In an examination, a candidate is required to answer 7 questions out of 12, which are divided into two groups, each containing 6 questions. One cannot attempt more than 5 questions from either group. In how many ways can he choose these questions? Solution: There are total 13 questions out of which 10 is to be answered .The student can answer in the following ways: $\Rightarrow 3$ questions from part $A$ and 4 from part $B$ $\Rightarrow 4$ questions from part $A$ and 3 from part $B$ $\...
Read More →In an examination, a student has to answer 10 questions,
Question: In an examination, 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 these questions be chosen? Solution: There are total 13 questions out of which 10 is to be answered .The student can answer in the following ways: $\Rightarrow 6$ questions from part $A$ and 4 from part $B$ $\Rightarrow 5$ questions from part $A$ and 5 from part $B$ $\Rightarrow 4$ questions from part $A$ a...
Read More →In an examination, a student has to answer 4 questions out of 5
Question: In an examination, a student has to answer 4 questions out of 5. Questions 1 and 2 are compulsory. Find the number of ways in which the student can make a choice. Solution: A student has to answer 4 questions out of 5 in which he is compelled to do the 1 and 2 questions compulsory. So he has to attempt 2 questions from 3 of his choice. Choosing 2 questions from 3 will be in ${ }^{3} \mathrm{C}_{2}$ ways. Applying ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ $=3$ ways....
Read More →A question paper has two parts, part A and part B, each containing
Question: A question paper has two parts, part A and part B, each containing 10 questions. If the student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions? Solution: The question paper has two sets each containing 10 questions. So the student has to choose 8 from part A and 5 from part B. $\Rightarrow$ choosing 8 questions from 10 of part $A$ in ${ }^{10} C_{8}$ $\Rightarrow$ choosing 5 questions from 10 of part $B$ in ${ }^{10} C_{5}$ $\Rightarrow$ by ...
Read More →In how many ways can a cricket team be selected from a group of 25 players
Question: In how many ways can a cricket team be selected from a group of 25 players containing 10 batsmen, 8 bowlers, 5 all-rounders and 2 wicketkeepers, assuming that the team of 11 players requires 5 batsmen, 3 all-rounders, 2 bowlers and 1 wicketkeeper? Solution: A team of 11 players is to be made from 25 players. $\Rightarrow$ Selecting 5 batsmen from 10 in ${ }^{10} \mathrm{C}_{5}$ ways. $\Rightarrow$ Selecting 3 all-rounders from 5 in ${ }^{5} \mathrm{C}_{3}$ ways. $\Rightarrow$ Selecting...
Read More →A cricket team of 11 players is to be selected from 16 players including
Question: A cricket team of 11 players is to be selected from 16 players including 5 bowlers and 2 wicketkeepers. In how many ways can a team be selected so as to consist of exactly 3 bowlers and 1 wicketkeeper? Solution: There is a cricket team of 11 players is to be selected from 16 players, which must include 3 bowlers and a wicketkeeper. $\Rightarrow$ There will be a team of 7 batsmen, 1 wicketkeeper and 3 bowlers. $\Rightarrow$ There are 5 bowlers from which 3 is to be selected in ${ }^{5} ...
Read More →From 4 officers and 8 clerks, in how many ways can 6 be chosen
Question: From 4 officers and 8 clerks, in how many ways can 6 be chosen (i) to include exactly one officer, (ii) to include at least one officer? Solution: The team of 6 has to be chosen from 4 officers and 8 clerks. There are some restrictions which are 1. To include exactly one officer In this case, One officer will be chosen from 4 in ${ }^{4} \mathrm{C}_{1}$ ways Therefore, 5 will be chosen form 8 clerks in ${ }^{8} \mathrm{C}_{5}$ ways. Thus by multiplication principle, we get Total no. of...
Read More →A sports team of 11 students is to be constituted, choosing at least 5
Question: 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 team be constituted? Solution: There are 20 students in each classes and there is need of at least 5 students in each class to form a team of team of 11. Now There are two ways in which the selection can be possible 1. Selecting 5 from $\mathrm{XI}$ and 6 from $\mathrm{XII}$ 2. Selecting 6 from $\...
Read More →In How many ways can a student chose 5 courses out of 9 courses
Question: In How many ways can a student chose 5 courses out of 9 courses if 2 specific courses are compulsory for every student? Solution: Since every student needs to choose 5 courses out of which 2 are compulsory. So, the student needs to choose 3 subjects out of 7. No. of ways for choosing 3 subjects out of 7 is ${ }^{7} \mathrm{C}_{3}$ Applying formula ${ }^{n} \mathrm{C}_{r}=\frac{n !}{r !(n-r) !}$ , we get = 35 ways....
Read More →How many different boat parties of 8 consisting of 5 boys and 3 girls can be
Question: How many different boat parties of 8 consisting of 5 boys and 3 girls can be made from 20 boys and 10 girls. Solution: Number of ways of choosing 5 boys out of 20 boys $={ }^{20} \mathrm{C}_{5}$ $=\frac{20 !}{5 ! \times(20-5) !}$ $=\frac{20 !}{5 ! \times 15 !}$ $=19 \times 17 \times 16 \times 3=15,504$ Number of ways of choosing 3 girls out of 10 girls $={ }^{10} \mathrm{C}_{3}$ $=\frac{10 !}{3 ! \times(10-3) !}$ $=\frac{10 !}{3 ! \times 7 !}$ $=15 \times 8=120$ Total number of ways $=...
Read More →Find the number of ways of selecting 9 balls from 6 red balls, 5 while balls
Question: Find the number of ways of selecting 9 balls from 6 red balls, 5 while balls and 4 blue balls if each selection consists of 3 balls of each colour. Solution: Total number of red balls = 6 Total number of white balls $=5$ Total number of blue balls $=4$ No. of ways of selecting 3 balls which is red $={ }^{6} \mathrm{C}_{3}$ No. of ways of selecting 3 balls which is white $={ }^{5} \mathrm{C}_{3}$ No. of ways of selecting 3 balls which is blue $={ }^{4} \mathrm{C}_{3}$ Thus, by Multiplic...
Read More →A bag contains 5 black and 6 red balls.
Question: A bag contains 5 black and 6 red balls. Find the number of ways in which 2 black and 3 red balls can be selected. Solution: There are 5 black and 6 red balls. So, The number of ways of selecting 2 black balls from 5 black balls is ${ }^{5} \mathrm{C} 2$, and number of ways of selecting 3 red balls from 6 red balls is ${ }^{6} C_{3}$. Thus using the multiplication principle, the total number of ways will be $\Rightarrow{ }^{5} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{3}$ ways. Applying ...
Read More →In how many ways can 5 sportsmen be selected from a group of 10?
Question: In how many ways can 5 sportsmen be selected from a group of 10? Solution: As there are 10 sportsmen out of which 5 are to be selected. 5 sportsmen can be selected out of 10 in ${ }^{10} \mathrm{C}_{5}$ ways. Applying ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ We get, $\Rightarrow{ }^{10} C_{5}=\frac{10 !}{5 !(10-5) !}$ $\Rightarrow 252$ ways Hence, there are 252 ways of selecting 5 sportsmen from 10 sportsmen....
Read More →From a class of 25 students, 4 are to be chosen for a competition.
Question: From a class of 25 students, 4 are to be chosen for a competition. In how many ways can this be done? Solution: This is a case of combination: Here $\mathrm{n}=25$ $r=4$ $\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{25} \mathrm{C}_{4}$ $\Rightarrow{ }^{25} \mathrm{C}_{4}=\frac{25 !}{(25-4) ! \times 4 !}$ $\Rightarrow{ }^{25} \mathrm{C}_{4}=\frac{25 !}{21 ! \times 4 !}$ $\Rightarrow{ }^{25} \mathrm{C}_{4}=\frac{25 \times 24 \times 23 \times 22 \times 21 !}{21 ! \times 4 !}$ ...
Read More →How many chords can be drawn through 21 points on a circle?
Question: How many chords can be drawn through 21 points on a circle? Solution: Number of points=21 ⇒n=21 A chord connects circle at two points. ⇒r=2 $\Rightarrow$ Number of chords from 21 points $={ }^{n} C_{r}$ $\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{21} \mathrm{C}_{2}$ $\Rightarrow{ }^{21} C_{2}=\frac{21 !}{(21-2) ! \times 2 !}$ $\Rightarrow{ }^{21} \mathrm{C}_{2}=\frac{21 !}{19 ! \times 2 !}$ $\Rightarrow{ }^{21} C_{2}=\frac{21 !}{19 ! \times 2 !}$ $\Rightarrow{ }^{21} C_{2...
Read More →If there are 12 persons in a party and if each two of them shake
Question: If there are 12 persons in a party and if each two of them shake hands with each other, how many handshakes are possible? Solution: With 12 people , we need to choose a subset of two different people where order does not matter. Also, we need to choose all such subsets because each person is shaking hands with everyone else exactly once. The number of ways is: ${ }^{n} \bar{C}_{r}$ Where: $n=12 \ r=2$ ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{12} \mathrm{C}_{2}$ $\Rightarrow{ }^{1...
Read More →How many different teams of 11 players can be chosen from 15 players?
Question: How many different teams of 11 players can be chosen from 15 players? Solution: Condition: Each student has an equal chance of getting selected. Imagine selecting the teammates one at a time. There are 15 ways of selecting the first teammate, 14 ways of selecting the second, 13 ways of selecting the third teammate, and so on down to 5 ways of selecting the eleventh teammate. This is a problem of combination $\Rightarrow \mathrm{n}=15 \ \mathrm{r}=11$ $\Rightarrow{ }^{n} C_{r}={ }^{15} ...
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Question: If ${ }^{n+1} C_{r+1}:{ }^{n} C_{r}=11: 6$ and ${ }^{n} C_{r}:{ }^{n-1} C_{r-1}=6: 3$, find $n$ and $r$. Solution: Given: ${ }^{n+1} C_{r+1}:{ }^{n} C_{r}=11: 6$ and ${ }^{n} C_{r}:{ }^{n-1} C_{r-1}=6: 3$ To Find : $n \ r$ We use this property in this question: $\left(\begin{array}{l}n \\ r\end{array}\right)=\frac{n}{r} \times\left(\begin{array}{l}n-1 \\ r-1\end{array}\right)$ ${ }^{n+1} C_{r+1}:{ }^{n} C_{r}=11: 6$ $\Rightarrow \frac{\left(\begin{array}{l}n+1 \\ r+1\end{array}\right)}...
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Question: If ${ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$, find $r .$ Solution: Given, ${ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$ To find:r=? $\frac{\left(\begin{array}{c}n \\ r\end{array}\right)}{\left(\begin{array}{c}n \\ r-1\end{array}\right)}=\frac{84}{36}=\frac{7}{3}$ $\frac{\frac{n !}{(n-r) ! \times r !}}{\frac{n !}{(n-r+1) ! \times(r-1) !}}=\frac{7}{3}$ $\Rightarrow \frac{(n-r+1)}{r}=\frac{7}{3}$ $\Rightarrow 3(n-r+1)=7 r$ $\Rightarrow 3 n-10 r=-3 \...
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Question: If ${ }^{n} P_{r}=840$ and ${ }^{n} C_{r}=35$, find the value of $r$. Solution: Given: ${ }^{n} P_{r}=840$ and ${ }^{n} C_{r}=35$ To find: $r=?$ We know that: ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{(n-r) ! \times r !}$ and $\mathrm{n} \mathrm{P}_{\mathrm{r}}=\frac{n !}{(n-r) !}$ $\Rightarrow{ }^{n} P_{r}={ }^{n} C_{r} \times r !$ $\Rightarrow 840=35 \times r !$ ⇒ $r !=\frac{840}{35}=24$ $\Rightarrow r !=4 !$ $\Rightarrow r=4$ Ans: $r=4$...
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Question: If ${ }^{15} C_{r}:{ }^{15} C_{r-1}=11: 5$, find $r$. Solution: Given: ${ }^{15} \mathrm{C}_{r}:{ }^{15} \mathrm{C}_{r-1}=11: 5$ To find: $r=?$ ${ }^{15} \mathrm{C}_{\mathrm{r}}:{ }^{15} \mathrm{C}_{\mathrm{r}-1}=11: 5$ $\frac{\left(\begin{array}{c}15 \\ r\end{array}\right)}{\left(\begin{array}{c}15 \\ r-1\end{array}\right)}=\frac{11}{5}$ $\frac{\frac{15 !}{(15-r) ! r !}}{\frac{15 !}{(16-r) !(r-1) !}}=\frac{11}{5}$ $\Rightarrow \frac{15 !(16-r) !(r-1) !}{(15-r) ! r ! 15 !}=\frac{11}{5}...
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Question: If ${ }^{2 n} C_{3}:{ }^{n} C_{3}=12: 1$, find $n .$ Solution: Given: ${ }^{2 n} C_{3}:{ }^{n} C_{3}=12: 1$ To find: $\mathrm{n}=$ ? ${ }^{2 n} \mathrm{C}_{3}:{ }^{n} \mathrm{C}_{3}=12: 1$ $\Rightarrow \frac{\left(\begin{array}{c}2 n \\ 3\end{array}\right)}{\left(\begin{array}{c}n \\ 3\end{array}\right)}=\frac{12}{1}$ $\frac{\frac{(2 n) !}{(2 n-3) ! 3 !}}{\frac{n !}{(n-3) ! 3 !}}=\frac{12}{1}$ $\Rightarrow \frac{(2 n) !(n-3) ! 3 !}{(2 n-3) ! 3 ! n !}=\frac{12}{1}$ $\Rightarrow \frac{(2...
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Question: If ${ }^{n} C_{r-1}={ }^{n} C_{3 r}$, find $r$. Solution: Given: ${ }^{n} \mathrm{C}_{r-1}={ }^{n} \mathrm{C}_{3 r}$ To find: $r=?$ We know that: ${ }^{n} C_{r}={ }^{n} C_{n-r}$ $\Rightarrow{ }^{n} C_{r-1}={ }^{n} C_{n-(r-1)}$ $\Rightarrow{ }^{n} C_{r-1}={ }^{n} C_{n-r+1}$ $\Rightarrow{ }^{n} C_{n-r+1}={ }^{n} C_{3 r}$ $\Rightarrow n-r+1=3 r$ $\Rightarrow 4 r=n+1$ ⇒ $r=\frac{n+1}{4}$ Ans: $r=\frac{n+1}{4}$...
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