Solve this
Question: If ${ }^{m} C_{1}={ }^{n} C_{2}$ prove that $m=\frac{1}{2} n(n-1)$ Solution: Given: ${ }^{m} C_{1}={ }^{n} C_{2}$ Need to prove: $m=\frac{1}{2} n(n-1){ }^{m} C_{1}={ }^{n} C_{2} \Rightarrow \frac{m !}{1 !(m-1) !}=\frac{n !}{2 !(n-2) !} \Rightarrow$ $\frac{m(m-1) !}{(m-1) !}=\frac{1}{2} \frac{n(n-1)(n-2) !}{(n-2) !} \Rightarrow m=\frac{1}{2} n(n-1)$ [Proved]...
Read More →Find the values of
Question: Find the values of (i) ${ }^{200} \mathrm{C}_{198}$, (ii) ${ }^{76} \mathrm{C}_{0}$, (iii) ${ }^{15} \mathrm{C}_{15}$ Solution: (i) ${ }^{200} C_{198}$ $=\frac{200 !}{198 !(200-198) !}=\frac{200 !}{198 ! 2 !}=\frac{200 \times 199}{2 !}=100 \times 199=19900$ (ii) ${ }^{76} \mathrm{C}_{0}$ $=\frac{76 !}{0 ! 76 !}=1_{[\text {As } 0 !=1]}$ (iii) ${ }^{15} \mathrm{C}_{15}$ $=\frac{15 !}{15 ! 0 !}=1$...
Read More →Solve this
Question: If ${ }^{35} C_{n+7}={ }^{35} C_{4 n-2}$ then find the value of $n$. Solution: Given: ${ }^{35} \mathrm{C}_{n+7}={ }^{35} \mathrm{C}_{4 n-2}$ Need to find: Value of $n$ We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t \mathrm{OR}$ (ii) $r+t=n$ Applying property (i) we get, $\Rightarrow \mathrm{n}+7=4 \mathrm{n}-2 \Rightarrow 3 \mathrm{n}=9 \Rightarrow \mathrm{n}=3$ Applying property (ii) we get, $\Rightarrow \mathrm{n}+7+4 \mathrm{n}-2=$ ...
Read More →Solve this
Question: If ${ }^{n} C_{r+1}={ }^{n} C_{8}$ then find the value of ${ }^{22} C_{n}$. Solution: Given: ${ }^{n} C_{r+1}={ }^{n} C_{8}$ Need to find: Value of ${ }^{22} C_{n}$ We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t \mathrm{OR}$ (ii) $r+t=n$ We are going to use property (i) ${ }^{n} C_{r+1}={ }^{n} C_{8}$ By the property (i), $\Rightarrow r+1=8 \Rightarrow r=7$ Now we are going to use property (ii) $\Rightarrow \mathrm{n}=8+7+1=16$ Therefor...
Read More →Solve this
Question: If ${ }^{20} C_{r+1}={ }^{20} C_{r-10}$ then find the value of ${ }^{10} C_{r}$. Solution: Given: ${ }^{20} \mathrm{C}_{r+1}={ }^{20} \mathrm{Cr}-10$ Need to find: Value of ${ }^{10} \mathrm{Cr}$ We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t$ OR (ii) $r+t=n$ We can't apply the property (i) here. So we are going to use property (ii) ${ }^{20} \mathrm{Cr}+1={ }^{20} \mathrm{Cr}-10 \mathrm{By}$ the property (ii), $\Rightarrow r+1+r-10=20 ...
Read More →Solve this
Question: If ${ }^{20} C_{r+1}={ }^{20} C_{r-10}$ then find the value of ${ }^{10} C_{r}$. Solution: Given: ${ }^{20} C_{r+1}={ }^{20} C_{r-10}$ Need to find: Value of ${ }^{10} C_{r}$ We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C t$, then, (i) $r=t$ OR (ii) $r+t=n$ We can't apply the property (i) here. So we are going to use property (ii) ${ }^{20} \mathrm{Cr}+1={ }^{20} \mathrm{Cr}-10$ By the property (ii), $\Rightarrow r+1+r-10=20 \Rightarrow 2 r=29 \Rightarrow r...
Read More →Solve this
Question: If ${ }^{20} C_{r}={ }^{20} C_{r-10}$ then find the value of ${ }^{17} C_{r}$. Solution: Given: ${ }^{20} \mathrm{C}_{r}={ }^{20} \mathrm{C}_{r-10}$ Need to find: Value of ${ }^{17} \mathrm{C}_{r}$ We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t$ OR (ii) $r+t=n$ We can't apply the property (i) here. So we are going to use property (ii) ${ }^{20} C_{r}={ }^{20} C_{r-10}$ By the property (ii), $\Rightarrow r+r-$ $10=20 \Rightarrow 2 r=30 \...
Read More →Solve this
Question: If ${ }^{20} C_{r}={ }^{20} C_{r-10}$ then find the value of ${ }^{17} C_{r}$. Solution: Given: ${ }^{20} \mathrm{C}_{r}={ }^{20} \mathrm{C}_{r-10}$ Need to find: Value of ${ }^{17} \mathrm{C}_{r}$ We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t \mathrm{OR}$ (ii) $r+t=n$ We can't apply the property (i) here. So we are going to use property (ii) ${ }^{20} C_{r}={ }^{20} C_{r-10}$ By the property (ii), $\Rightarrow r+r-$ $10=20 \Rightarrow...
Read More →How many words, with or without meaning, can be formed from the letters of
Question: How many words, with or without meaning, can be formed from the letters of the word, MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time? (ii) All letters are used at a time? (iii) All letters are used, but the first letter is a vowel? Solution: There are 6 letters in the word MONDAY, and there is no letter repeating. (i) 4 letters are used at first. 4 letters can sit in different ways. So, here permutation is to be used. So, the number of words that can be...
Read More →In how many ways can 4 girls and 3 boys be seated in a row so that no two boys are together?
Question: In how many ways can 4 girls and 3 boys be seated in a row so that no two boys are together? Solution: The seating arrangement would be like this: B G B G B G B G B So, 4 girls can seat among the four places. Number of ways they can seat is $={ }^{4} \mathrm{P}_{4}=24$ Boys have to seat among the ' $B$ ' areas. So, there are 5 seats available for 3 boys. The number of ways the 3 boys can seat among the 5 places is $={ }^{5} \mathrm{P}_{3}=60$ Therefore, the total number of ways they ca...
Read More →The English alphabet has 21 consonants and 5 vowels. How many words
Question: The English alphabet has 21 consonants and 5 vowels. How many words with two different consonants and three different vowels can be formed from the alphabet? Solution: 2 consonants out of 21 consonants can be chosen in ${ }^{21} C_{2}$ ways. 3 vowels out of 5 vowels can be chosen in ${ }^{5} C_{3}$ ways. Length of the word is $=(2+3)=5$ And also 5 letters can be written in 5! Ways. Therefore, the number of words can be formed is = $\left({ }^{21} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}...
Read More →The English alphabet has 21 consonants and 5 vowels. How many words
Question: The English alphabet has 21 consonants and 5 vowels. How many words with two different consonants and three different vowels can be formed from the alphabet? Solution: 2 consonants out of 21 consonants can be chosen in ${ }^{21} C_{2}$ ways. 3 vowels out of 5 vowels can be chosen in ${ }^{5} C_{3}$ ways. Length of the word is $=(2+3)=5$ And also 5...
Read More →How many words, each of 3 vowels and 2 consonants
Question: How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word INVOLUTE? Solution: In the word INVOLUTE there are 4 vowels, I,O,U and E and there are 4 consonants, 'N','V','L' and 'T'. 3 vowels out of 4 vowels can be chosen in ${ }^{4} \mathrm{C}_{3}$ ways. 2 consonants out of 4 consonants can be chosen in ${ }^{4} \mathrm{C}_{2}$ ways. Length of the formed words will be $(3+2)=5 .$ So, the 5 letters can be written in $5 !$ Ways. Therefore, the total numb...
Read More →Out of 12 consonants and 5 vowels, how many words, each containing 3
Question: Out of 12 consonants and 5 vowels, how many words, each containing 3 consonants and 2 vowels, can be formed? Solution: 3 consonants out of 12 consonants can be chosen in ${ }^{12} C_{3}$ ways. 2 vowels out of 5 vowels can be chosen in ${ }^{5} \mathrm{C}_{2}$ ways. And also 5 letters can be written in 5 ! Ways. Therefore, the number of words can be formed is $\left({ }^{12} C_{3} X^{5} C_{2} X 5 !\right)=264000$....
Read More →There are 18 points in a plane of which 5 are collinear.
Question: There are 18 points in a plane of which 5 are collinear. How many straight lines can be formed by joining them? Solution: A line is formed by joining two points. If the total number of points is 18 , the total number of lines would be $={ }^{18} \mathrm{C}_{2}$ But 5 points are collinear, so the lines made by these points are the same and would be only $1 .$ Hence there is 1 common line joining the 5 collinear points. As these 5 points are also included in 18 points so these must be su...
Read More →Find the number of ways in which a committee of 2 teachers and 3 students
Question: Find the number of ways in which a committee of 2 teachers and 3 students can be formed out of 10 teachers and 20 students. In how many of these committees (i) a particular teacher is included? (ii) a particular student is included? (iii) a particular student is excluded? Solution: Since a committee is to be formed of 2 teachers and 3 students (i) When a particular teacher is included No. of ways in which committee can be formed $={ }^{9} \mathrm{C}_{1} \times{ }^{20} \mathrm{C}_{3}$ =...
Read More →How many different products can be obtained by multiplying two or more
Question: How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 without repetition? Solution: The given no. is 3,5,7,11. The no. of different products when two or more is taking= the no. of ways of taking the product of two no.+ the no. of ways of taking the product of three no. + the no. of ways of taking the product of four no. $={ }^{4} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{4}$ Applying ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$ $=6+...
Read More →How many different selections of 4 books can be made from 10 different
Question: How many different selections of 4 books can be made from 10 different books, if (i) there is no restriction? (ii) two particular books are always selected? (iii) two particular books are never selected? Solution: Since there are 10 different books out of which 4 is to be selected . (i) When there is no restriction No. of ways in which 4 books be selected $={ }^{10} \mathrm{C}_{4}$ = 210 ways (ii) Two particular books are always selected Since two particular books are always selected, ...
Read More →How many triangles can be formed in a decagon?
Question: How many triangles can be formed in a decagon? Solution: Total number of sides in a decagon = 10 We know that number of vertices in triangle $=3$ So, out of 10 vertices we have to choose 3 vertices. Therefore, Total number of triangles in a decagon $={ }^{10} \mathrm{C}_{3}$ $=\frac{10 !}{3 ! \times(10-3) !}=\frac{10 \times 9 \times 9}{3 \times 2}$ Total number of triangles = 120...
Read More →How many triangles can be obtained by joining 12 points, four
Question: How many triangles can be obtained by joining 12 points, four of which are collinear? Solution: Total number of points on plane = 12 Triangles can be formed from these points $={ }^{12} \mathrm{C}_{3}$ = 220 But 4 points are colinear, the number of triangles can be formed from these points $={ }^{4} \mathrm{C}_{3}$ = 4 We need to subtract 4 from 220 because in the formation of triangles from 4 colinear points are added there. So no of triangle formed is $=220-4$ = 216...
Read More →Find the number of diagonals of
Question: Find the number of diagonals of (i) a hexagon, (ii) a decagon, (iii) a polygon of 18 sides Solution: For a diagonal to be formed, 2 vertices are required. Thus in a polygon, there are 10 sides. And no. of lines can be formed are ${ }^{n} C_{2}$, but in ${ }^{n} C_{2}$ the sides are also included. N of them is sides. Thus the no. of diagonals are ${ }^{n} C_{2}-n$ (i) Hexagon $\mathrm{N}=6$ so no of diagonal is ${ }^{6} \mathrm{C}_{2}-6$ $=9$ (ii) decagon $N=10$ So no of diagonal is ${ ...
Read More →Find the number of 5-card combinations out of a deck of 52 cards if a least
Question: Find the number of 5-card combinations out of a deck of 52 cards if a least one of the five cards has to be king. Solution: Since there are 52 cards in a deck out of which 4 are king and others are nonkings. So, the no. of ways are as follows: 1. 1 king and 4 non-king 2. 2 king and 3 non-king 3. 3 king and 2 non-king 4. 4 king and 1 non-king So, total no. of ways are $={ }^{4} C_{1} \times{ }^{48} C_{4}+{ }^{4} C_{2} \times{ }^{48} C_{3}+{ }^{4} C_{3} \times{ }^{48} C_{2}+{ }^{4} C_{4}...
Read More →From a class of 14 boys and 10 girls, 10 students are to be chosen for a
Question: From a class of 14 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? Solution: 2 girls who won the prize last year are surely to be taken. So, we have to make a selection of 8 students out of 14 boys and 8 girls, choosing at least 4 boys and at least 2 girls. Thus, we may choose: (4 boys, 4 girls) or ( 5 boys, 3 girls) or ( 6 b...
Read More →A committee of 5 is to be formed out of 6 gents and 4 ladies.
Question: A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done, when (i) at least 2 ladies are included? (ii) at most 2 ladies are included? Solution: Since the committee of 5 is to be formed from 6 gents and 4 ladies. (i) Forming a committee with at least 2 ladies Here the possibilities are (i) 2 ladies and 3 gents (ii) 3 ladies and 2 gents (iii) 4 ladies and 1 gent The number of ways they can be selected $={ }^{4} \mathrm{C}_{2}$ $\times$ ${ }^{6} \ma...
Read More →A committee of three persons is to be constituted from a group
Question: A committee of three 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? Solution: Total number of persons = 2 + 3 = 5 Now, committee consist of 3 persons. Therefore, total number of ways $={ }^{5} \mathrm{C}_{3}$ $=\frac{5 !}{3 ! \times(5-3) !}$ $=5 \times 2=10$ Now, When 1 man is selected, total ways $={ }^{2} \mathrm{C}_{1}$ When 2 women are selected, total ways $={ }^{3} ...
Read More →