Mathmetics

ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams

Q511.
How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI' using each letter exactly once ?
1) 60 2) 120
3) 180 4) 240
5)None of these
Answer : 120
Explanation :
The word 'DELHI' has 5 letters and all these letters are different.

Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once

= Number of arrangements of 5 letters taken all at a time =  5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120
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Q512.
In how many ways can the letters of the word 'LEADER' be arranged ?
1) 60 2) 120
3) 240 4) 360
5)None of these
Answer : 360
Explanation :
The word 'LEADER' has 6 letters. 

But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters = 6!2!=6×5×4×3×2×12×1 = 360
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Q513.
There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed ?

1) 8750 2) 10420
3) 11760 4) 12420
5)None of these
Answer : 11760
Explanation :
We need to select 5 men from 8 men and 6 women from 10 women

Number of ways to do this  = 8C5 x 10C6

=  8C3 x 10C4 [Applied the formula nCr = nC(n - r) ]

= (8×7×63×2×1) (10×9×8×74×3×2×1)

= 56 x 210 = 11760
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Q514.
Out of 7 consonants and 4 vowels, how many words of 3  consonants and 2 vowels can be formed ?

1) 2100 2) 24400
3) 21300 4) 25200
5)None of these
Answer : 25200
Explanation :
Number of ways of selecting 3 consonants out of 7 = 7C3

Number of ways of selecting 2 vowels out of 4 = 4C2

Number of ways of selecting 3 consonants out of 7 and 2

vowels out of 4 = 7C3 x 4C2

=(7×6×5/3×2×1)×(4×3/2×1)=210

It means that we can have 210 groups where each group

contains total 5 letters(3 consonants and 2 vowels).

Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120

Hence, Required number of ways = 210 x 120 = 25200

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Q515.
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together ?
1) 100800 2) 120960
3) 240150 4) 4989600
5)None of thesae
Answer : 120960
Explanation :
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8! / (2!)(2!)= 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4! /2! = 12.
So, required number of words = (10080 x 12) = 120960.

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Q516.
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women ?
1) 42 2) 48
3) 63 4) 90
5)None of these
Answer : 63
Explanation : The required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = (7 x 6 )/(2 x 1) x 3 =  63.
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Q517.
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 ?
1) 24 2) 32
3) 36 4) 48
5)None of these
Answer : 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.

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Q518.
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated ?
1) 6 2) 10
3) 12 4) 20
5)None of thse
Answer : 20
Explanation :
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
So, required number of numbers = (1 x 5 x 4) = 20.
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Q519.
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together ?
1) 810 2) 1440
3) 2880 4) 50400
5)57600
Answer : 50400
Explanation :
In the word  'CORPORATION', we consider the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3!= 20 ways.
So, required number of ways = (2520 x 20) = 50400.
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Q520.
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together ?
1) 360 2) 480
3) 720 4) 1280
5)None of these
Answer : 720
Explanation :
The word 'LEADING'' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

 So, required number of ways = (120 x 6) = 720.
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