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# Solved Examples(Set 1) - Permutation and Combination

 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

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
$=210×120=25200$

 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

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)]

$=\dfrac{6×5}{2×1}+\dfrac{6×5×4}{3×2×1}×4$ $+\dfrac{6×5}{2×1}×\dfrac{4×3}{2×1}+6×4$
$=15+80+90+24=209$
 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

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) ]

$= \dfrac{7×6}{2×1}+\dfrac{7×6×5}{3×2×1}×6$ $+\dfrac{7×6×5}{3×2×1}×\dfrac{6×5}{2×1}$
$=21+210+525\\=756$
 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

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

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? 0 0 reply Jay
2015-06-24 09:05:59
You are looking for integer partitions. Refer Counting Integral Solutions 0 0 Kushal Vyas
2015-06-15 08:20:13
why o!=1??? 0 0 reply 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. 0 0 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. 0 0 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? 0 0 reply 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 0 0 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 0 0 sam
2016-01-07 11:48:21
"it will be the same as vowels comes together."

No. With your logic, QEUAIOTN is also a valid arrangement which is not. Vowels and consonants must be side by side, like 'EUAIO'QTN 0 0 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 0 0
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