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