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A box contains 4 red, 3 white, and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls in different colors
168
12
24
48
Answer with explanation
Answer: Option CExplanation
This question seems to be a bit typical, isn’t, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24
Please note that we have multiplied the combination results, we use to add when there is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.
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How many 3 digit numbers are divisible by 4?
125
158
225
252
Answer with explanation
Answer: Option CExplanation
A number is divisible by 4 when its last two digits are divisible by 4
For this, the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, an = a + (n-1)d
====>>>> 96 = 0 + (n-1)*4
==== >>>> n = 25
so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
====>>>> so total 9*25 = 225
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Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.
310
320
330
340
Answer with explanation
Answer: Option CExplanation
Here the order of choosing the elements doesn’t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
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Groups each containing 3 boys are to be formed out of 5 boys. A, B, C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?
8
6
5
7
Answer with explanation
Answer: Option DExplanation
Maximum number of such different groups = ABC, ABD, ABE, BCE,BDE,CEA,DEA =7.
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There are 5 floating stones on a river. A man wants to cross the river. He can move either 1 or 2 steps at a time. Find the number of ways in which he can cross the river?
13
12
14
11
Answer with explanation
Answer: Option AExplanation
The man needs to take 6 steps to cross the river. He can do this in the following ways:
===>>> Crossing the river by 6 unit steps = 1 way.
===>>> Crossing the river by 4 unit steps and 1 double step = 5C1 = 5C4 = 5 ways.
===>>> Crossing the river by 2 unit steps and 2 double steps = 4C2 = 6 ways.
===>>> Crossing the river by 3 double steps = 1 way.
Hence, the required number of ways = 1 + 5 + 6 + 1 = 13.
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How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
1000
100
50
25
Answer with explanation
Answer: Option BExplanation
1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ‘n’.
=> Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.
=> Note distinct initials is different from initials where the digits are different.
=> For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
=>This n3 different initials = 1 million
======>>>>> i.e. n3 = 106 (1 million = 106)
======>>>>> n3 = (102)3 => n = 102 = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
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Nine chairs are numbered 1 to 9. Three women and four men wish to occupy one chair each. First, the women chose the chairs from amongst the chair marked 1 to 5; and then the men select the chairs from amongst the remaining. The number of possible arrangements is
5C2 × 4P3
5C3 × 6C4
5C3 × 4C2
None of these
Answer with explanation
Answer: Option BExplanation
Women can select 3 chairs from chairs numbered 1 to 5 in 5C3 ways and remaining 6 chairs can be selected by 4 men in 6C4 ways. Hence the required number of ways = 5C3 × 6C4
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How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?
360
720
1420
1680
Answer with explanation
Answer: Option DExplanation
The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.
Given that no digit appears more than once. Hence we have 8 digits remaining (0,1,2,4,6,7,8,9)(0,1,2,4,6,7,8,9)
So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways.
Total number of ways = 8P4 =8×7×6×5=1680
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How many different four letter words can be formed (the words need not be meaningful using the letters of the word “MEDITERRANEAN” such that the first letter is E and the last letter is R?
53
55
59
69
Answer with explanation
Answer: Option CExplanation
The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or have both the letters to be the same.
Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.
Total number of possibilities = 56 + 3 = 59
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In how many ways can a committee consisting of three men and four women be formed from a group of six men and seven women?
⁶C₃ ⁷C₅
⁶C₃⁷C₄
⁷C₅ ⁶C₄
⁶C₃ ⁷C₃
Answer with explanation
Answer: Option BExplanation
The group contained six men and seven women
Three men can be selected from six men in ⁶C₃ ways.
Four women can be selected from seven women in ⁷C₄ ways.
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A box contains 2 red coins, 3 green coins and 4 blue coins. In how many ways can 3 coins be chosen such that at least one coin is green?
128
32
16
64
Answer with explanation
Answer: Option DExplanation
There are three cases:
Therefore, total number of ways = 3C3 + 3C2 * 6C1 + 3C1 * 6C2 = 1 + 3*6 + 3*15 = 64.
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