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401.

A man has 8children to take them to a zoo. He takes thre of them at a time to the zoo as often as he can without taking the same 3 children together more than once. How many times will he have to go to the zoo? How many times a particular child will to to the zoo?A. 56B. 21C. 112D. none of these

Answer» Correct Answer - B
Discussion
402.

A father with 8 children takes 3 at a time to the Zoological Gardens, as often as he can without taking the same 3 children together more than once. The number of times he will go to the garden, isA. 336B. 112C. 56D. none of these

Answer» Correct Answer - C
Discussion
403.

A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.

Answer» Since, condidate cannot attempt more than 5 questions from either group.
Thus, he is able to attempt minimum two questions from either group.
The number of questions attempted from each group is given in following table.
`{:("Group I ", 5, 4, 3, 2),("Group II ", 2, 3, 4, 5):}`
Since, each group have 6 questions and total attempted 7 questions.
`therefore" "` Total number of possible ways `= ""^(6)C_(5) xx ""^(6)C_(2)+ ""^(6)C_(4) xx ""^(6)C_(3) + ""^(6)C_(3) xx ""^(6)C_(4) + ""^(6)C_(2) xx ""^(6)C_(5)`
`= 2[""^(6)C_(5) xx ""^(6)C_(2) + ""^(6)C_(4) xx ""^(6)C_(3)]`
`= 2 [6 xx 15 + 15 xx 20]`
`= 2 [90 + 300]`
`= 2 xx 390 = 780`
Discussion
404.

Prove that `.^(n-1)C_(3)+.^(n-1)C_(4) gt .^(n)C_(3)` if `n gt 7`.

Answer» We have, `.^(n-1)C_(3)+.^(n-1)C_(4) gt .^(n)C_(3) " "[because.^(n)C_(r)+.^(n)C_(r-1)=.^(n+1)C_(r)]`
`hArr .^(n)C_(4)gt.^(n)C_(3)`
`hArr (n!)/(4!(n-4)!)gt(n!)/(3!(n-3)!)`
`hArr (1)/(4(n-4)!)gt(1)/((n-3)(n-4)!)" "[becausem!=m(m-1)!]`
`hArr n-3 gt 4 hArr n gt7`
Discussion
405.

Number of positive terms in the sequence `x_n=195/(4P_n)-(n+3p_3)/(P_(n+1)), n in N` (here `p_n=|anglen`)

Answer» Correct Answer - 4
We have, `x_(n)=(195)/(4P_(n))-(.^(n+3)A_(3))/(P_(n+1))`
`thereforex_(n)=(195)/(4*n!)-((n+3)(n+2)(n+1))/((n+1)!)`
`=-(195)/(4*n!)-((n+3)(n+2))/(n!)`
`=(195-4n^(2)-20n-24)/(4*n!)=(171-4n^(2)-20n)/(4*n!)`
`becausex_(n)` is positive
`therefore(171-4n^(2)-20n)/(4*n!) gt0`
`implies4n^(2)+20n-171 lt0`
which is true for n=1,2,3,4
Hence, the given sequence `(x_(n))` has 4 positive terms.
Discussion
406.

(i)In how manyways can a pack of 52 cards be divided equally among four players?(ii)(ii) In howmany ways can you divide these cards in four sets, three of them having 17cards each and the four the one just one card?

Answer» Correct Answer - `(i) ((52)!)/((13!)^(4))" " (ii) ((52)!)/(4!(13!)^(4)) " " (iii) ((52)!)/(3!(17)^(3))`
(i) The number of ways in which a pack of 52 cards can be divided equally into four groups of 13 cards
each `= ""^(52)C_(13) xx ""^(39)C_(13) xx ""^(26)C_(13) xx ""^(13)C_(13) = ((52)!)/((13!)^(4))`
(ii) The number of ways in which a pack of 52 cards can be divided equally into four groups of 13 cards
each `= (""^(52)C_(13) xx ""^(39)C_(13) xx ""^(26)C_(13) xx ""^(13)C_(13))/(4!) = ((52)!)/(4!(13!)^(4))`
(iii) The number of ways in which a pack of 52 cards be divided into 4 sets, three of them having 17 cards each and the fourth just one card
`= (""^(52)C_(17) xx ""^(35)C_(18) xx ""^(18)C_(17) xx ""^(1)C_(1))/(3!) = ((52)!)/(3!(17!)^(4))`
Discussion
407.

The number of ways in which we can distribute `m n`studentsequally among `m`sections isgiven bya. `((m n !))/(n !)`b. `((m n)!)/((n !)6m)`c. `((m n)!)/(m ! n !)`d. `(m n^m)`A. `((mn)!)/(n!)`B. `((mn)!)/((n!)^(m))`C. `((mn)!)/(m!n!)`D. `((2n)!)/((n!)^(2))`

Answer» Here, the ordering of sections is important.
So, required number of ways `=(((mn)!)/((n!)^(m)m!))m!""=((mn)!)/((n!)^(m))`
Discussion
408.

The number of possible outcomes when a coin is tossed 6 times isA. 36B. 64C. 12D. 32

Answer» Correct Answer - B
Number of outcomes when tossing a coin 1 times = 2 (head or tail)
`therefore` Total possible outcomes when a coin tossed 6 times `= 2^(6) = 64` [`because 2^(n)` for n time tossed coin]
Discussion
409.

Three dice are rolled. Find the number of possible outcomes in which atleast one dice shows 5.A. 215B. 36C. 125D. 91

Answer» Correct Answer - D
Discussion
410.

The numbers of four different digits number that can be formed from the digits of the number 12356 such that the numbers are divisible by 4, isA. 36B. 48C. 12D. 24

Answer» Correct Answer - A
Discussion
411.

If ` ^n C_(12)= ^n C_8`then `n=`A. 20B. 12C. 6D. 30

Answer» Correct Answer - A
Given that, `""^(n)C_(12) = ""^(n)C_(8)`
`rArr` `""^(n)C_(n - 12) = ""^(n)C_(8) " " [because ""^(n)C_(r) = ""^(n)C_(n-r)]`
`rArr` `n - 12 = 8`
`rArr` `n = 12 + 8 = 20`
Discussion
412.

The total number of five-digit numbers ofdifferent digits in which the digit in the middle is the largest isa. `sum_(n=4)^9^n P_4`b. `33(3!)`c. `30(3!)`d.none of theseA. `30xx30!`B. `33xx3!`C. `sum_(n=4)^(9)""^(n)C_(4)xx4!`D. none of these

Answer» The largest digit in the middle of the five digit number can be 4 (when numbers are formed with digits 0, 1, 2, 3, 4), 5 (when 0, 1, 2, 3, 4, 5 are used),…..,9 (when 0, 1, 2,……,9 are used).
`:.` Number of five digit numbers with largest digit 5 in the middle
`=(""^(4)C_(4)xx4!-""^(3)C_(3)xx3!)`
Number of five digit numbers with lardest digit 5 in the middle
`=(""^(5)C_(4)xx4!-""^(3)C_(3)xx3!)`
Hence, required number of numbers
`=(""^(4)C_(4)xx4!-""^(3)C_(3)xx3!)+(""^(5)C_(4)xx4!-""^(4)C_(3)xx3!)`
`+(""^(6)C_(4)xx4!-""^(5)C_(3)xx3!)+.....+(""^(9)C_(4)+4!-""^(8)C_(3)xx3!)`
Discussion
413.

The number of 5-digit number that can be made using the digits 1 and 2 and in which at laest one digits is different, isA. 30B. 31C. 32D. none of these

Answer» We have,
Required number of numbers
= Total numbres of 5-gigit numbers formed by the digits 1 and 2
- Number of 5-digit numbers formed by the digits 1 and 2 when numbers have all digits equal
`=2^(5)-2=30`.
Discussion
414.

A set contains 2n+1 elements. The number of subsets of this set containing more than n elements :A. `2^(n-1)`B. `2^(n)`C. `2^(n+1)`D. `2^(2n)`

Answer» Correct Answer - D
Discussion
415.

Find number of seating arrangements of 6 persons at three identical round tables if every table must be the occupied

Answer» Total possible ways`=(6*5)/2*3+(6*5)/2*4+(6*5)/2*(4*3)/2`
`=45+60+90`
`=195`.
Discussion
416.

In a city no two persons have identical set of teeth and there is no peson without a tooth. Also no person has more than 32 teeth. If we disguard the shape and size of tooth and consider only the positioning of the teeth, the maximum population of the city isA. `2^(32)`B. `(32)^(2)-1`C. `2^(32)-1`D. `2^(31)`

Answer» Correct Answer - C
Discussion
417.

The number of ways in which 5 picturers can be hung from 7 picture nails on the wall isA. `7^(5)`B. `5^(7)`C. 2520D. none of these

Answer» Correct Answer - C
Discussion