1. In how many ways a four digit even number can be formed by using the digits 2,3,5,8 exactly once?
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Correct Ans:12
Explanation:
Solution is
Given
Four digit even number can be formed by using the digits 2,3,5,8
Since the number has to be an even number,the unit digit has to be 2 or 8.
First three places can be filled by remaining three digits.
= 3! + 3!
&nb
2. In a plane there are totally 8 points (no three points are collinear), how many lines can be drawn ?
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Correct Ans:28
Explanation:
To draw a line we need 2 points.
From the given 8 points , we need to choose 2 points
Number of ways in which this can be done is 18C2= (8 x 7 ) / 2 = 28
3. How many 4 digit even number can formed by using the digits 1,3,7 and 8 only once ?
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Correct Ans:6
Explanation:
We can form 24 numbers with digits 1,3,7 and 8 using only once.
To be an even number, the units digit has to even number, in this case only 8 should come in the unit's place . The remaining places, 3 positions can be filled in 3! , that is 6 ways
4. How many words can be formed with or without meaning by taking all the letters from the word TAKEN ?
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Correct Ans:120
Explanation:
There are totally 4 letters in the word TAKEN
T, A, K, E and N.
No. Of Words = 5 x 4 x 3 x 2 x 1 = 120
5. In how many different ways can the letters of the word 'OFFICES' be arranged ?
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Correct Ans:2520
Explanation:
The word OFFICES contains O,F, F I, C, E, S contains 7 letters, in which 2 are identical.
No. Of words that can formed = 7 ! / 2 ! = 2520
SHOW ANSWER
Correct Ans:600
Explanation:
25! / 23! = 23! X 24 x 25 / ( 23!) = 24 x 25 = 600
7. In a party there were totally 20 people, each person shook his hands with the other person. How many hand shakes would have taken place ?
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Correct Ans:190
Explanation:
There are 20 people. Every person has to shake hands with the other person. Which means we have to find the number of ways of choosing 2 people from the 20.
The number of ways it ca happen = 20C2 = ( 20 x 19 ) / (1 x 2) = 190
8. A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting a 4 digit code with the proper combination of each of the 4 rings Maximum how many codes can be formed to open the lock?
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Correct Ans:9 x 9 x 9 x 9
Explanation:
9 x 9 x 9 x 9 = 94
9. How many words can be formed with or without meaning by taking all the letters from the word SMALL ?
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Correct Ans:60
Explanation:
Number of letters in SMALL is 5, but there are 2 L?s
Answer is 5! / 2! = 120 / 2 = 60
10. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
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Correct Ans:720
Explanation:
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways
= (120 x 6)
= 720.
11. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
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Correct Ans:5040
Explanation:
LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= ^{10}P_{4} = (10 x 9 x 8 x 7) = 5040.
12. In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
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Correct Ans:36
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under: (1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
13. In how many ways can the letters of the word 'LEADER' be arranged ?
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Correct Ans:360
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Therefore, Required number of ways = 6! / (1!)(2!)(1!)(1!)(1!) = 360
14. How many numbers greater than a million can be formed by using the digits 7, 4, 6 and 0 if 4 has to be used twice, 6 has to be used thrice and the rest only once?
SHOW ANSWER
Correct Ans:360
Explanation:
The given digits are 4, 4, 6, 6, 6, 7 and 0. Totally we have 7 digits. So numbers greater than a million can be formed by using all the digits.
4 occurs twice, 6 occurs thrice while 0 and 7 once,
Therefore total number of arrangements = 7! / (3! x 2!) = 420
We have to avoid 0 in the starting place, Number of ways in which 0 can be in the first place is 6! / 2! X 3! = 60
Hence the total number of ways = 420 – 60 = 360
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