Permutations and Combinations

Three flags each of different colours are available for a military exercise, Using these flags different codes can be generated by waving I. Single flag of different colours II. Any two flags in a different sequence of colours. III. three flags in a different sequence of colours. The maximum number of codes that can be generated is.

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Three flags each of different colours are available for a military exercise, Using these flags different codes can be generated by waving
I. Single flag of different colours
II. Any two flags in a different sequence of colours.
III. three flags in a different sequence of colours.
The maximum number of codes that can be generated is.

A. 6
B. 9
C. 15
D. 18
Explanation:
This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).
We can choose any colour.
Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.
Hence total number of ways by changing flag = 3+ 6 +6 = 15

A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?

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A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?

A. 63
B. 511
C. 1023
D. 15
Explanation:
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 – 1.
= 210 – 1 = 1024 – 1 = 1023

In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?

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In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?

A. 25
B. 13
C. 40
D. 30
Explanation:
The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

How many four digit even numbers can be formed using the digits {2, 3, 5, 1, 7, 9}

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How many four digit even numbers can be formed using the digits {2, 3, 5, 1, 7, 9}

A. 60
B. 360
C. 120
D. 240
Explanation:
The given digits are 1, 2, 3, 5, 7, 9
A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only ‘2’ and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60.

The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is_________?

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The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is_________?

A. (6!)2
B. 6! * ⁷P₆
C. 2(6!)
D. 6! * 7
Explanation:
We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways = 6! * ⁷P₆

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most__________?

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A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most__________?

A. 216
B. 243
C. 215
D. 729
Explanation:
Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 – 1 = 215.

A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?

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A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?

A. 165
B. 185
C. 205
D. 225
Explanation:
Total number of persons in the committee = 5 + 6 = 11
Number of ways of selecting group of eight persons = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 ways.