Permutation and Combination questions for Placement

Solved Examples(Set 1) - Permutation and Combination

answer with explanation

Answer: Option A

Explanation:

Number of ways of selecting 3 consonants from 7
= 7C3
Number of ways of selecting 2 vowels from 4
= 4C2

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
$=\left(\dfrac{7 × 6 × 5}{3 × 2 × 1}\right) × \left(\dfrac{4 × 3}{2 × 1}\right) \\= 210$

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
$=210×120=25200$

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 = 6C4

We can select 3 boys and 1 girl ...(option 2)
Number of ways to this = 6C3 × 4C1

We can select 2 boys and 2 girls ...(option 3)
Number of ways to this = 6C2 × 4C2

We can select 1 boy and 3 girls ...(option 4)
Number of ways to this = 6C1 × 4C3

Total number of ways
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1[∵ nCr = nC(n-r)]

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

$=15+80+90+24=209$

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 = 7C5

We can select 4 men and 1 woman ...(option 2)
Number of ways to do this = 7C4 × 6C1

We can select 3 men and 2 women ...(option 3)
Number of ways to do this = 7C3 × 6C2

Total number of ways
= 7C5 + (7C4 × 6C1) + (7C3 × 6C2)
= 7C2 + (7C3 × 6C1) + (7C3 × 6C2)[∵ nCr = nC(n - r) ]

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

$=21+210+525\\=756$

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
$=120×6=720$

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?

Jay

2015-06-24 09:05:59 

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.

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
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.

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

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

sam

2016-01-07 11:48:21 

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

No. With your logic, Q

TN is also a valid arrangement which is not. Vowels and consonants must be side by side, like '

'QTN

Alia's answer is correct.

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

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