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The ways of selecting 4 letters from the word EXAMINATION is
- a)136
- b)130
- c)125
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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The ways of selecting 4 letters from the word EXAMINATION isa)136b)130...
To find the ways of selecting 4 letters from the word EXAMINATION, we can use the combination formula:
nCr = n! / r!(n-r)!
Where n is the total number of items, r is the number of items to be selected, and ! denotes factorial.
So, in this case:
n = 11 (total number of letters in EXAMINATION)
r = 4 (number of letters to be selected)
Using the formula, we get:
11C4 = 11! / 4!(11-4)!
= (11 x 10 x 9 x 8) / (4 x 3 x 2 x 1)
= 11 x 5 x 3 x 4 x 2
= 1320
However, this includes combinations that contain repeated letters (e.g. selecting both Es or both Is). To exclude these combinations, we need to subtract the number of combinations that contain repeated letters.
There are two Es, two Is, and one each of the other letters. So, the number of combinations with repeated letters can be found by adding the number of combinations with two Es, the number of combinations with two Is, and the number of combinations with one each of the other letters:
2C2 x 9C2 + 2C2 x 7C2 + 7C1 x 4C1 x 3C2 = 36 + 21 + 63 = 120
Therefore, the total number of combinations without repeated letters is:
1320 - 120 = 1200
Hence, the correct option is A, i.e. 136.
nCr = n! / r!(n-r)!
Where n is the total number of items, r is the number of items to be selected, and ! denotes factorial.
So, in this case:
n = 11 (total number of letters in EXAMINATION)
r = 4 (number of letters to be selected)
Using the formula, we get:
11C4 = 11! / 4!(11-4)!
= (11 x 10 x 9 x 8) / (4 x 3 x 2 x 1)
= 11 x 5 x 3 x 4 x 2
= 1320
However, this includes combinations that contain repeated letters (e.g. selecting both Es or both Is). To exclude these combinations, we need to subtract the number of combinations that contain repeated letters.
There are two Es, two Is, and one each of the other letters. So, the number of combinations with repeated letters can be found by adding the number of combinations with two Es, the number of combinations with two Is, and the number of combinations with one each of the other letters:
2C2 x 9C2 + 2C2 x 7C2 + 7C1 x 4C1 x 3C2 = 36 + 21 + 63 = 120
Therefore, the total number of combinations without repeated letters is:
1320 - 120 = 1200
Hence, the correct option is A, i.e. 136.
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The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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