InterviewSolution

Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.	If `.^(n)P_(5)=20 .^(n)P_(3)`, find the value of n.
Answer» Correct Answer - 5040 `(n!)/((n-5)!)xx((n-3)!)/(n!)=20` or (n-3)(n-4)=20 or n=-1,8 But -1 is not acceptable.

2.	How many 4-letter words, with or without meaning, can be formed out of the letters in the word LOGARITHMS, if repetition of letters is not allowed ?
Answer» Correct Answer - 7 There are 10 letters in the word LOGARITHMS. So, the number of 4-letter words is equal to the number of arrangements of 10 letters, taken 4 at a time, i.e., `.^(10)P_(4)=5040`.

3.	How any three digit odd numbers can be formed by using the digits1,2,3,4,5,6 if;The repetition of digits is not allowed?The repetition of digits is allowed?A. 60B. 108C. 120D. 216
Answer» Correct Answer - B Total no. of digits = 6 To form a odd numbers we have only 3 choice for the unit digits. Now, Extreme left place can be filled in 6 ways the middle place can be filled in 6 ways. `:.` Required number of numbers `= 6 xx 6 xx 3 = 108`

4.	Find the number of ways of selection of at least one vowel and oneconsonant from the word TRIlPLE.
Answer» We have consonants T, R, P, L and vowels I, E. Number of ways of selecting at least one consonant `=2^(4)-1=15` Number of ways of selecting at least one vowel `=2^(2)-1=3` So, required number of selections `=15xx3=45`

5.	Eight chairs are numbered 1 to 8. Two women andthree men wish to occupy one chair each. First, the women choose the chairsfrom amongst the chairs marked 1 to 4, and then the men select th chairs fromamongst the remaining. The number of possible arrangements isa.`^6C_3xx^4C_2`b. `^4P_2xx^4P_3`c. `^4C_2xx^4P_3`d. none of these
Answer» Correct Answer - 468000 First women choose the chairs from amongst the chairs 1 to 4. Two women can be arranged in 4 chairs in `.^(4)P_(2)` ways. In remaining 6 chairs 3 men can be arranged in `.^(6)P_(3)` ways. `therefore` Total number of possible arrangements `= .^(4)P_(2)xx .^(6)P_(3)` `=(4!)/(2!)xx(6!)/(3!)` ltbgt `=4xx3xx6xx5xx4` =1440.

6.	Find the three-digit odd numbers that can be formed by using the digits1, 2, 3, 4, 5, 6 when the repetition is allowed.
Answer» Correct Answer - 108 Extreme left place can be filled in 6 ways and the extreme right place in only 3 ways. Since the number to be number to be formed is odd, so the required number is `6xx6xx3=108`.

7.	A tea party is arranged for 2m people along two sides of a long table with m chairs on each side, r men wish to sit on one particular side and s on the other. IN how many ways can they be seates ? `[r,s,lem]`A. `24 xx 8! xx 8!`B. `(81)^(3)`C. `210 xx 8! xx 8!`D. `16!`
Answer» Correct Answer - C Number of ways `= (8!9!)/(4!6!) xx 10!` `=(8!)^(2) xx (10!)/(4!6!)` `= (8!)^(2) xx (10 xx 9 xx 8 xx 7)/(4!)` `= (8!)^(2) xx (10 xx 9 xx 8 xx 7)/(4 xx 3 xx 2 xx1)` `= (210) xx (8!)^(2)`

8.	How many numbers between 100 an 1000 can be formed with the digits 5, 6, 7, 8, 9, if the repetition of digits is not allowed ?A. `3^(5)`B. `5^(3)`C. 120D. 60
Answer» Correct Answer - D Number between 100 and 1000 are 3-digit numbers. It is given that the digits should not be repeated. Number of given digits = 5 In a 3-digit number, first number can be arranged in 5 ways. Second number in 4 ways. Third number is 3 ways `:.` Numbers that can be formed `= 5 xx 4 xx 3 = 60`

9.	The number of triangles that can be formed by choosing the vefrom a set of 12 points, seven of which lie on the same straight line, are:A. 185B. 175C. 115D. 105
Answer» Correct Answer - A To form a triangle, we need 3 points. 12 points are given. So, `.^(12)C_(3)` triangle can be formed ltbr. But, givne that 7 points are on a straight line. Selecting 3 points from this set will not form a triangle So, number of triangle formed `.^(12)C_(3) - .^(7)C_(3)` `= (12!)/(3!9!) - (7!)/(3!4!)` `= (12 xx 11 xx 10 )/(3 xx 2 xx 1) - (7 xx 6 xx 5)/(3 xx 2 xx 1) = 220 - 35 = 185`

10.	Find the number of non-negative integral solutions of equation `x+y+z+2w=20.`
Answer» If w=0, equation reduces to x+y+z=20. Number of non-negative inteegral solutions `= .^(20+3-1)C_(3-1) = .^(22)C_(2)` If w=1, equation reduces to x+y+z=18. Number of non-negative integral solutions `=(18+3-1)C_(3-1)= .^(20)C_(2)` Similarly, we get number of non-negative integral solutions for w=2, 3, .., 10. Hence, total number of solutions `=(22)C_(2)+ .^(20)C_(2)+ .^(18)C_(2)+..+ .^(2)C_(2)`

11.	Find the number of non-negative integral solutions of `x+y+z+wlt=20.`
Answer» We have inequality, `x+y+z+w le 20` This is equivalent to equation x+y+z+w+t=20, where `t ge 0` Clearly number of non-negative integral solutions to (1) and (2) are same. So, number of solutions `= .^(20+5-1)C_(5-1)= .^(24)C_(4)`

12.	How many four-digit numbers divisible by 10 can be formed using 1, 5, 0, 6, 7 without repetition of digits ?A. 24B. 36C. 44D. 64
Answer» Correct Answer - A A number divisible by 10 means that last digit is 0 So, the remaining 3 digits can be arranged in `4 xx 3 xx 2` ways = 24 ways.

13.	Find the number of positive integral solutions satisfying the equation `(x_1+x_2+x_3)(y_1+y_2)=77.`
Answer» Correct Answer - 420 `(x_(1)+x_(2)+x_(3))(y_(1)+y_(2))=11xx7 " or " 7xx11` In the first case, `(x_(1)+x_(2)+x_(3))=11 " and" (y_(1)+y_(2))=7`, which have ` .^(10)C_(2). .^(6)C_(1)` solutions In the second case, `(x_(1)+x_(2)+x_(3))=7 " and" (y_(1)+y_(2))=11`, which have `.^(6)C_(2). .^(10)C_(1)` solutions `therefore` Total number of solutions `= .^(10)C_(2). .^(6)C_(1)+ .^(6)C_(2). .^(10)C_(1)` =270+150=420

14.	Find the number of non-negative integral solutions of the equation `x+y+z=10.`
Answer» Here the number of solutions is equivalent to number of ways 10 identical objects can be distributed among three persons if empty groups are allowed that is `.^(10+3-1)C_(3-1)= .^(12)C_(2)=66`.