1. In how many ways a four digit even number can be formed by using the digits 2,3,5,8 exactly once?
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!
2. In a plane there are totally 8 points (no three points are collinear), how many lines can be drawn ?
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 ?
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 ?
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 ?
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
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 ?
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?
Correct Ans:9 x 9 x 9 x 9
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 ?
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?
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)
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?
LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P4 = (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?
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 ?
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?
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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