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1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? | |

A. 25200 | B. 21300 |

C. 24400 | D. 210 |

Discuss |

answer with explanation

Answer: Option A

Explanation:

Number of ways of selecting 3 consonants from 7

= ^{7}C_{3}

Number of ways of selecting 2 vowels from 4

= ^{4}C_{2}

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4

= ^{7}C_{3} × ^{4}C_{2}

$=\left(\dfrac{7 × 6 × 5}{3 × 2 × 1}\right) × \left(\dfrac{4 × 3}{2 × 1}\right) \\= 210$

It means 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

$=5!=5×4×3×2×1=120$

Hence, required number of ways

$=210×120=25200$

2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? | |

A. 212 | B. 209 |

C. 159 | D. 201 |

Discuss |

answer with explanation

Answer: Option B

Explanation:

In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there.

Hence we have 4 options as given below

We can select 4 boys ...(option 1)

Number of ways to this = ^{6}C_{4}

We can select 3 boys and 1 girl ...(option 2)

Number of ways to this = ^{6}C_{3} × ^{4}C_{1}

We can select 2 boys and 2 girls ...(option 3)

Number of ways to this = ^{6}C_{2} × ^{4}C_{2}

We can select 1 boy and 3 girls ...(option 4)

Number of ways to this = ^{6}C_{1} × ^{4}C_{3}

Total number of ways

= ^{6}C_{4} + ^{6}C_{3} × ^{4}C_{1} + ^{6}C_{2} × ^{4}C_{2} + ^{6}C_{1} × ^{4}C_{3}

= ^{6}C_{2} + ^{6}C_{3} × ^{4}C_{1} + ^{6}C_{2} × ^{4}C_{2} + ^{6}C_{1} × ^{4}C_{1}[∵ ^{n}C_{r} = ^{n}C_{(n-r)}]

$=\dfrac{6×5}{2×1}+\dfrac{6×5×4}{3×2×1}×4$ $+\dfrac{6×5}{2×1}×\dfrac{4×3}{2×1}+6×4$

$=15+80+90+24=209$3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. In how many ways can it be done? | |

A. 702 | B. 624 |

C. 756 | D. 812 |

Discuss |

answer with explanation

Answer: Option C

Explanation:

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.

Hence we have the following 3 options.

We can select 5 men ...(option 1)

Number of ways to do this = ^{7}C_{5}

We can select 4 men and 1 woman ...(option 2)

Number of ways to do this = ^{7}C_{4} × ^{6}C_{1}

We can select 3 men and 2 women ...(option 3)

Number of ways to do this = ^{7}C_{3} × ^{6}C_{2}

Total number of ways

= ^{7}C_{5} + (^{7}C_{4} × ^{6}C_{1}) + (^{7}C_{3} × ^{6}C_{2})

= ^{7}C_{2} + (^{7}C_{3} × ^{6}C_{1}) + (^{7}C_{3} × ^{6}C_{2})[∵ ^{n}C_{r} = ^{n}C_{(n - r) }]

$= \dfrac{7×6}{2×1}+\dfrac{7×6×5}{3×2×1}×6$ $+\dfrac{7×6×5}{3×2×1}×\dfrac{6×5}{2×1}$

$=21+210+525\\=756$4. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? | |

A. 920 | B. 825 |

C. 720 | D. 610 |

Discuss |

answer with explanation

Answer: Option C

Explanation:

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).

Hence we can assume total letters as 5 and all these letters are different.

Number of ways to arrange these letters

$=5!=5×4×3×2×1=120$

All the 3 vowels (OIA) are different

Number of ways to arrange these vowels among themselves

$=3!=3×2×1=6$

Hence, required number of ways

$=120×6=720$

5. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? | |

A. 42000 | B. 48000 |

C. 50400 | D. 47200 |

Discuss |

answer with explanation

Answer: Option C

Explanation:

The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN(OOAIO).

Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and rest of the letters are different.

Number of ways to arrange these letters

$=\dfrac{7!}{2!}=\dfrac{7×6×5×4×3×2×1}{2×1}=2520$

In the 5 vowels (OOAIO), 'O' occurs 3 and rest of the vowels are different.

Number of ways to arrange these vowels among themselves $=\dfrac{5!}{3!}=\dfrac{5×4×3×2×1}{3×2×1}=20$

Hence, required number of ways

$=2520×20=50400$

vijay

2015-06-24 01:56:35

in how many ways sum s can be formed using exactly n variables......

input

5 -->(s)

2---->(n)

output

4

reason :

2+3 3+2 1+4 4+1

can anyone help me to find the logic?

input

5 -->(s)

2---->(n)

output

4

reason :

2+3 3+2 1+4 4+1

can anyone help me to find the logic?

Jay

2015-06-24 09:05:59

You are looking for integer partitions. Refer Counting Integral Solutions

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shuvam gupta

2015-06-23 06:44:35

There are three proof:

A)Pattern

4!=5!/5

3!=4!/4

2!=3!/3

1!=2!/2

And

0!=1!/1===1...

B)Practical sums:

No.of ways of arranging

The word"APPLE"

ANS

as there are 2 p and 1 a,l,e

So,5!/(2!*1!*1!*1!)

But other24 alphabets are not there so they are 0 in no.

For them it will be 0!

If 0!=1*0=0

So our answer must be undefined,but in reality it is not

So ,0!=1

C)Practical thinking:

If there are 3 items we can arrange in 3! Ways

If there are 2 items we can arrange in 2! Ways

If there are 1 items we can arrange in1! Ways

If there are 0 items we can arrange in 1 way...it present state...so0!=1.

A)Pattern

4!=5!/5

3!=4!/4

2!=3!/3

1!=2!/2

And

0!=1!/1===1...

B)Practical sums:

No.of ways of arranging

The word"APPLE"

ANS

as there are 2 p and 1 a,l,e

So,5!/(2!*1!*1!*1!)

But other24 alphabets are not there so they are 0 in no.

For them it will be 0!

If 0!=1*0=0

So our answer must be undefined,but in reality it is not

So ,0!=1

C)Practical thinking:

If there are 3 items we can arrange in 3! Ways

If there are 2 items we can arrange in 2! Ways

If there are 1 items we can arrange in1! Ways

If there are 0 items we can arrange in 1 way...it present state...so0!=1.

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Manu

2015-07-30 04:56:07

Hai,Bro what u said is right and simply we can learn by C programming for 0! is easy way of understanding. Don't think bad its just an example for knowing the factorial case. Those who know C language it is easily understandable..

Ex:

#include<stdio.h>

main()

{

int fact=1,i,num;

printf("Enter the Number:\n");

scanf("%d",&num);

for(i=1;i<=num;i++)

{

fact=fact*i;

}

printf("%d\n",fact);

}

It can be understood simply that. The factorial of 0!=1 is always 1. If not the factorial case is wrong.

Ex:

#include<stdio.h>

main()

{

int fact=1,i,num;

printf("Enter the Number:\n");

scanf("%d",&num);

for(i=1;i<=num;i++)

{

fact=fact*i;

}

printf("%d\n",fact);

}

It can be understood simply that. The factorial of 0!=1 is always 1. If not the factorial case is wrong.

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Andrea

2015-04-04 05:42:18

How many words with or without dictionary meaning can be formed using the letters of the word EQUATION so the vowel and consonant are side by side?

alia

2015-09-16 11:51:51

there are 5 vowels so 5! ways of arranging them.

3 consonants, so 3! ways of arranging them.

overall there are 2 ways to arrange the groups of vowels and consonants.

Hence the answer would be 2! x 5!x 3! = 1440

3 consonants, so 3! ways of arranging them.

overall there are 2 ways to arrange the groups of vowels and consonants.

Hence the answer would be 2! x 5!x 3! = 1440

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ajay gupta

2016-01-07 11:47:18

no ..it will be the same as vowels comes together.. means 3 consonants and one group of vowels as asumed 4 and such that 4! ways and 5 vowels are arranged in 5! ways and hence = 4!*5!= 2880 ways

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sam

2016-01-07 11:48:21

"it will be the same as vowels comes together."

No. With your logic, Q__EUAIO__TN is also a valid arrangement which is not. Vowels and consonants must be side by side, like '__EUAIO__'QTN

Alia's answer is correct.

No. With your logic, Q

Alia's answer is correct.

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Dev

2015-04-06 16:22:57

Assuming that all letters needs to be used.

EQUATION - 5 vowels and 3 consonants

5 vowels - can be arranged in 5! ways

3 consonants - can be arranged in 3! ways

each of this group can be arranged in 2! ways

Total number of ways = 5! * 3! * 2! = 1440

Required number of words = 1440

EQUATION - 5 vowels and 3 consonants

5 vowels - can be arranged in 5! ways

3 consonants - can be arranged in 3! ways

each of this group can be arranged in 2! ways

Total number of ways = 5! * 3! * 2! = 1440

Required number of words = 1440

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