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- Question
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
Options- A. 48
- B. 64
- C. 63
- D. 45
- Correct Answer
- 64
ExplanationWe may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways= = (45 + 18 + 1) =64
- 1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Options- A. 564
- B. 735
- C. 756
- D. 657 Discuss
- 2. In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?
Options- A. 2800
- B. 2880
- C. 2600
- D. 3980 Discuss
- 3. A box contains 5 green, 4 yellow and 3 white marbles. Threemarbles are drawn at random. What is the probability thatthey are not of the same colour ?
Options- A. 40/44
- B. 44/41
- C. 41/44
- D. 40/39 Discuss
- 4. In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?
Options- A. 1,51,200 ways.
- B. 5,04,020 ways
- C. 72,000 ways
- D. None of the above Discuss
- 5. In how many ways can the letters of the word 'LEADER' be arranged ?
Options- A. 360
- B. 420
- C. 576
- D. 220 Discuss
- 6. A standard deck of playing cards has 13 spades. How many ways can these 13 spades be arranged?
Options- A. 13!
- B. 13^2
- C. 13^13
- D. 2! Discuss
- 7. A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible ways of presenting these 10 cups to the expert.
Options- A. 340
- B. 210
- C. 290
- D. 252 Discuss
- 8. There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card ?
Options- A. 4 x 3^4
- B. 3^4
- C. 4^3
- D. 3 x 4^3 Discuss
- 9. Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?
Options- A. 3
- B. 6
- C. 4
- D. 2 Discuss
- 10. In how many ways can 100 soldiers be divided into 4 squads of 10, 20, 30, 40 respectively?
Options- A. 1700
- B. 18!
- C. 190
- D. None of these Discuss
Permutation and Combination problems
Search Results
Correct Answer: 756
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways= = 756.
Correct Answer: 2880
Explanation:
They have to be arranged in the following way :
L T L T L T L T L
The 5 lions should be arranged in the 5 places marked ?L?.
This can be done in 5! ways.
The 4 tigers should be in the 4 places marked ?T?.
This can be done in 4! ways.
Therefore, the lions and the tigers can be arranged in 5!´ 4! ways = 2880 ways.
Correct Answer: 41/44
Explanation:
n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220
Correct Answer: 1,51,200 ways.
Explanation:
In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E
Total = 13 letters
But last letter must be N
Hence, available places = 12
In that odd places = 1, 3, 5, 7, 9, 11
Owvels = 4
This can be done in 6P4 ways
Remaining 7 letters can be arranged in 7!/3! x 2! ways
Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.
Correct Answer: 360
Explanation:
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360
Correct Answer: 13!
Explanation:
The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.
13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800
Correct Answer: 252
Explanation:
Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252
Correct Answer: 4 x 3^4
Explanation:
The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).
For each value on the card the remainder can have 3 possible values.
The total number of possible sequences is: 4*3^4
Correct Answer: 3
Explanation:
Let the number of Rose plants be ?a?.
Let number of marigold plants be ?b?.
Let the number of Sunflower plants be ?c?.
20a+5b+1c=1000; a+b+c=100
Solving the above two equations by eliminating c,
19a+4b=900
b = (900-19a)/4
b = 225 - 19a/4----------(1)
b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)
Substituting (1) in (2),
0 < 225 - 19a/4 < 99
225 < -19a/4 < (99 -225)
=> 4 x 225 > 19a > 126 x 4
=> 900/19 > a > 505
a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.
Hence possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.
Correct Answer: None of these
Explanation:
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