\displaystyle \textbf{Question 1. }\text{If }{}^{20}C_r={} ^{20}C_{r-10},\text{ then }{}^{18}C_r\text{ is equal to}
\displaystyle \text{(a) }4896\qquad \text{(b) }816\qquad \text{(c) }1632\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle {}^{20}C_r={} ^{20}C_{r-10}
\displaystyle \therefore r+(r-10)=20
\displaystyle 2r=30
\displaystyle r=15
\displaystyle \therefore {}^{18}C_r={} ^{18}C_{15}={} ^{18}C_3=816
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If }{}^{20}C_r={} ^{20}C_{r+4},\text{ then }{}^rC_3\text{ is equal to}
\displaystyle \text{(a) }54\qquad \text{(b) }56\qquad \text{(c) }58\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle {}^{20}C_r={} ^{20}C_{r+4}
\displaystyle \therefore r+(r+4)=20
\displaystyle 2r=16
\displaystyle r=8
\displaystyle \therefore {}^rC_3={}^8C_3=56
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\text{If }{}^{15}C_{3r}={} ^{15}C_{r+3},\text{ then }r\text{ is equal to}
\displaystyle \text{(a) }5\qquad \text{(b) }4\qquad \text{(c) }3\qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle {}^{15}C_{3r}={} ^{15}C_{r+3}
\displaystyle \therefore 3r=r+3\text{ or }3r+r+3=15
\displaystyle r=\frac{3}{2}\text{ or }4r=12
\displaystyle \therefore r=3
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{If }{}^{20}C_{r+1}={} ^{20}C_{r-1},\text{ then }r\text{ is equal to}
\displaystyle \text{(a) }10\qquad \text{(b) }11\qquad \text{(c) }19\qquad \text{(d) }12
\displaystyle \text{Answer:}
\displaystyle {}^{20}C_{r+1}={} ^{20}C_{r-1}
\displaystyle \therefore (r+1)+(r-1)=20
\displaystyle 2r=20
\displaystyle r=10
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 5. }\text{If }C(n,12)=C(m,8),\text{ then }C(22,n)\text{ is equal to}
\displaystyle \text{(a) }231\qquad \text{(b) }210\qquad \text{(c) }252\qquad \text{(d) }303
\displaystyle \text{Answer:}
\displaystyle C(n,12)=C(m,8)
\displaystyle \text{Since the values are equal, }n=m\text{ and }12+8=n
\displaystyle \therefore n=20
\displaystyle C(22,n)=C(22,20)=C(22,2)
\displaystyle =\frac{22\times21}{2}=231
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 6. }\text{If }{}^mC_1={}^nC_2,\text{ then}
\displaystyle \text{(a) }2m=n\qquad \text{(b) }2m=n(n+1)\qquad \\ \text{(c) }2m=n(n-1)\qquad \text{(d) }2n=m(m-1)
\displaystyle \text{Answer:}
\displaystyle {}^mC_1={}^nC_2
\displaystyle m=\frac{n(n-1)}{2}
\displaystyle \therefore 2m=n(n-1)
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 7. }\text{If }{}^nC_{12}={}^nC_8,\text{ then }n=
\displaystyle \text{(a) }20\qquad \text{(b) }12\qquad \text{(c) }6\qquad \text{(d) }30
\displaystyle \text{Answer:}
\displaystyle {}^nC_{12}={}^nC_8
\displaystyle \therefore 12+8=n
\displaystyle n=20
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 8. }\text{If }{}^nC_r+{}^nC_{r+1}={} ^{n+1}C_x,\text{ then }x=
\displaystyle \text{(a) }r\qquad \text{(b) }r-1\qquad \text{(c) }n\qquad \text{(d) }r+1
\displaystyle \text{Answer:}
\displaystyle {}^nC_r+{}^nC_{r+1}={} ^{n+1}C_{r+1}
\displaystyle \therefore x=r+1
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 9. }\text{If }{}^{a^2-a}C_2={} ^{a^2-a}C_4,\text{ then }a=
\displaystyle \text{(a) }2\qquad \text{(b) }3\qquad \text{(c) }4\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle {}^{a^2-a}C_2={} ^{a^2-a}C_4
\displaystyle \therefore 2+4=a^2-a
\displaystyle a^2-a=6
\displaystyle a^2-a-6=0
\displaystyle (a-3)(a+2)=0
\displaystyle \therefore a=3
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 10. }{}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5\text{ is equal to}
\displaystyle \text{(a) }30\qquad \text{(b) }31\qquad \text{(c) }32\qquad \text{(d) }33
\displaystyle \text{Answer:}
\displaystyle {}^5C_0+{}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5=2^5
\displaystyle {}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5=2^5-{}^5C_0
\displaystyle =32-1=31
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 11. }\text{Total number of words formed by }2\text{ vowels and }3
\displaystyle \text{ consonants taken from} 4\text{ vowels and }5\text{ consonants is equal to}
\displaystyle \text{(a) }60\qquad \text{(b) }120\qquad \text{(c) }7200\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Choose }2\text{ vowels from }4\text{ vowels and }3\text{ consonants from }5\text{ consonants.}
\displaystyle \text{Number of selections}={}^4C_2\times{}^5C_3
\displaystyle \text{The }5\text{ selected letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number}={}^4C_2\times{}^5C_3\times5!
\displaystyle =6\times10\times120=7200
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 12. }\text{There are }12\text{ points in a plane. The number of straight lines}
\displaystyle \text{joining any two of them when }3\text{ of them are collinear, is}
\displaystyle \text{(a) }62\qquad \text{(b) }63\qquad \text{(c) }64\qquad \text{(d) }65
\displaystyle \text{Answer:}
\displaystyle \text{If no three points are collinear, number of lines}={} ^{12}C_2=66
\displaystyle \text{But }3\text{ collinear points give }1\text{ line instead of }{}^3C_2=3\text{ lines.}
\displaystyle \therefore \text{Required number}=66-3+1=64
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 13. }\text{Three persons enter a railway compartment. If there are }5\text{ seats vacant,}
\displaystyle \text{in how many ways can they take these seats?}
\displaystyle \text{(a) }60\qquad \text{(b) }20\qquad \text{(c) }15\qquad \text{(d) }125
\displaystyle \text{Answer:}
\displaystyle \text{Three persons can take }3\text{ seats out of }5\text{ seats in }{}^5P_3\text{ ways.}
\displaystyle {}^5P_3=5\times4\times3=60
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 14. }\text{In how many ways can a committee of }5\text{ be made out of }
\displaystyle 6\text{ men and }4\text{ women} \text{containing at least one women?}
\displaystyle \text{(a) }246\qquad \text{(b) }222\qquad \text{(c) }186\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total committees of }5\text{ from }10\text{ persons}={} ^{10}C_5=252
\displaystyle \text{Committees with no woman}={}^6C_5=6
\displaystyle \therefore \text{Required number}=252-6=246
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 15. }\text{There are }10\text{ points in a plane and }4\text{ of them are collinear.}
\displaystyle \text{The number of straight lines joining any two of them is}
\displaystyle \text{(a) }45\qquad \text{(b) }40\qquad \text{(c) }39\qquad \text{(d) }38
\displaystyle \text{Answer:}
\displaystyle \text{If no three points are collinear, number of lines}={} ^{10}C_2=45
\displaystyle \text{But }4\text{ collinear points give }1\text{ line instead of }{}^4C_2=6\text{ lines.}
\displaystyle \therefore \text{Required number}=45-6+1=40
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 16. }\text{There are }13\text{ players of cricket, out of which }4\text{ are bowlers. In how many}
\displaystyle \text{ways a team of eleven be selected from them so as to include at least two bowlers?}
\displaystyle \text{(a) }72\qquad \text{(b) }78\qquad \text{(c) }42\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ bowlers and }9\text{ other players.}
\displaystyle \text{A team of }11\text{ players excludes only }2\text{ players.}
\displaystyle \text{To have at least }2\text{ bowlers, the team may have }2,3\text{ or }4\text{ bowlers.}
\displaystyle \text{Required number}={}^4C_2{}^9C_9+{}^4C_3{}^9C_8+{}^4C_4{}^9C_7
\displaystyle =6\times1+4\times9+1\times36=78
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 17. }\text{If }C_0+C_1+C_2+\cdots+C_n=256,\text{ then }{}^{2n}C_2\text{ is equal to}
\displaystyle \text{(a) }56\qquad \text{(b) }120\qquad \text{(c) }28\qquad \text{(d) }91
\displaystyle \text{Answer:}
\displaystyle C_0+C_1+C_2+\cdots+C_n=2^n
\displaystyle 2^n=256=2^8
\displaystyle \therefore n=8
\displaystyle {}^{2n}C_2={} ^{16}C_2=\frac{16\times15}{2}=120
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 18. }\text{The number of ways in which a host lady can invite for a party of }
\displaystyle 8\text{ out of }12\text{ people } \text{of whom two do not want to attend the party together is}
\displaystyle \text{(a) }2\times{}^{11}C_7+{}^{10}C_8\qquad \text{(b) }{}^{10}C_8+{}^{11}C_7\qquad \text{(c) }{}^{12}C_8-{}^{10}C_6\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways of inviting }8\text{ people from }12={} ^{12}C_8
\displaystyle \text{Ways in which the two particular people are together}={} ^{10}C_6
\displaystyle \therefore \text{Required number}={} ^{12}C_8-{}^{10}C_6
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 19. }\text{Given }11\text{ points, of which }5\text{ lie on one circle, other than these }5,\text{ no }4
\displaystyle \text{lie on one circle. Then the number of circles that can be drawn so that each contains} \\ \text{at least }3\text{ of the given points is}
\displaystyle \text{(a) }216\qquad \text{(b) }156\qquad \text{(c) }172\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{If no }4\text{ points are concyclic, number of circles}={} ^{11}C_3=165
\displaystyle \text{The }5\text{ concyclic points give }1\text{ circle instead of }{}^5C_3=10\text{ circles.}
\displaystyle \therefore \text{Required number}=165-10+1=156
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 20. }\text{How many different committees of }5\text{ can be formed from }6\text{ men and }
\displaystyle 4\text{ women } \text{on which exact }3\text{ men and }2\text{ women serve?}
\displaystyle \text{(a) }6\qquad \text{(b) }20\qquad \text{(c) }60\qquad \text{(d) }120
\displaystyle \text{Answer:}
\displaystyle \text{Choose }3\text{ men from }6\text{ and }2\text{ women from }4.
\displaystyle \text{Required number}={}^6C_3\times{}^4C_2
\displaystyle =20\times6=120
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 21. }\text{If }{}^{43}C_{r-6}={} ^{43}C_{3r+1},\text{ then the value of }r\text{ is}
\displaystyle \text{(a) }12\qquad \text{(b) }8\qquad \text{(c) }6\qquad \text{(d) }10\qquad \text{(e) }14
\displaystyle \text{Answer:}
\displaystyle {}^{43}C_{r-6}={} ^{43}C_{3r+1}
\displaystyle \therefore r-6=3r+1\text{ or }(r-6)+(3r+1)=43
\displaystyle r=-\frac{7}{2}\text{ or }4r-5=43
\displaystyle \therefore r=12
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 22. }\text{The number of diagonals that can be drawn by joining the vertices} \\ \text{of an octagon is}
\displaystyle \text{(a) }20\qquad \text{(b) }28\qquad \text{(c) }8\qquad \text{(d) }16
\displaystyle \text{Answer:}
\displaystyle \text{Number of diagonals of an }n\text{-sided polygon}=\frac{n(n-3)}{2}
\displaystyle \text{For an octagon, }n=8
\displaystyle \therefore \text{Number of diagonals}=\frac{8(8-3)}{2}=20
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 23. }\text{The value of }\left({}^7C_0+{}^7C_1\right)+\left({}^7C_1+{}^7C_2\right)+\cdots+\left({}^7C_6+{}^7C_7\right)\text{ is}
\displaystyle \text{(a) }2^7-1\qquad \text{(b) }2^8-2\qquad \text{(c) }2^8-1\qquad \text{(d) }2^8
\displaystyle \text{Answer:}
\displaystyle \left({}^7C_0+{}^7C_1\right)+\left({}^7C_1+{}^7C_2\right)+\cdots+\left({}^7C_6+{}^7C_7\right)
\displaystyle ={}^7C_0+2({}^7C_1+{}^7C_2+\cdots+{}^7C_6)+{}^7C_7
\displaystyle =1+2(2^7-2)+1
\displaystyle =2^8-2
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 24. }\text{Among }14\text{ players, }5\text{ are bowlers. In how many ways a team of }11
\displaystyle \text{may be formed with at least }4\text{ bowlers?}
\displaystyle \text{(a) }265\qquad \text{(b) }263\qquad \text{(c) }264\qquad \text{(d) }275
\displaystyle \text{Answer:}
\displaystyle \text{There are }5\text{ bowlers and }9\text{ other players.}
\displaystyle \text{At least }4\text{ bowlers means }4\text{ bowlers or }5\text{ bowlers.}
\displaystyle \text{Required number}={}^5C_4{}^9C_7+{}^5C_5{}^9C_6
\displaystyle =5\times36+1\times84=264
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 25. }\text{A lady gives a dinner party for six guests. The number of ways in which}
\displaystyle \text{they may be selected from among ten friends if two of the friends will not} \\ \text{attend the party together is}
\displaystyle \text{(a) }112\qquad \text{(b) }140\qquad \text{(c) }164\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total ways of selecting }6\text{ guests from }10={} ^{10}C_6=210
\displaystyle \text{Ways in which the two particular friends are selected together}={} ^8C_4=70
\displaystyle \therefore \text{Required number}=210-70=140
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 26. }\text{If }{}^{n+1}C_3=2\cdot{}^nC_2,\text{ then }n=
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }5\qquad \text{(d) }6
\displaystyle \text{Answer:}
\displaystyle {}^{n+1}C_3=2\cdot{}^nC_2
\displaystyle \frac{(n+1)n(n-1)}{3!}=2\cdot\frac{n(n-1)}{2!}
\displaystyle \frac{n+1}{6}=1
\displaystyle n+1=6
\displaystyle n=5
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 27. }\text{The number of parallelograms that can be formed from a}
\displaystyle \text{set of four parallel lines intersecting another set of three parallel lines is}
\displaystyle \text{(a) }6\qquad \text{(b) }9\qquad \text{(c) }12\qquad \text{(d) }18
\displaystyle \text{Answer:}
\displaystyle \text{Choose }2\text{ lines from }4\text{ parallel lines and }2\text{ lines from }3\text{ parallel lines.}
\displaystyle \text{Required number}={}^4C_2\times{}^3C_2
\displaystyle =6\times3=18
\displaystyle \therefore \text{Correct option is (d).}
\\

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