Permutation and combinations quiz 1

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
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!=120
Hence, required number of ways
210*120=25200

Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = (6C1*4C3)+(6C2*4C2)+(6C3*4C1)+6C4

= (6C1*4C1)+(6C2*4C2)+(6C3*4C1)+6C2
= 209.

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)[∵ ⁿC_r = ⁿC_{(n - r)} ] $=\frac{7\times 6}{2\times 1}+\frac{7\times 6\times 5}{3\times 2\times 1}\times 6$ $+\frac{7\times 6\times 5}{3\times 2\times 1}\times \frac{6\times 5}{2\times 1}$ $=21+210+525=756$

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 $=5!=5\times 4\times 3\times 2\times 1=120$

All the 3 vowels (OIA) are different Number of ways to arrange these vowels among themselves $=3!=3\times 2\times 1=6$ Hence, required number of ways $=120\times 6=720$

Explanation: The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','Od' 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 $=\frac{7!/}{2!}=\frac{7\times 6\times 5\times 4\times 3\times 2\times \mathrm{1/}}{2\times 1}=2520$

In the 5 vowels (OOAIO), \'O\' occurs 3 and rest of the vowels are different. Number of ways to arrange these vowels among themselves $=\frac{5!/}{3!}=\frac{5\times 4\times 3\times 2\times \mathrm{1/}}{3\times 2\times 1}=20$

Hence, required number of ways $=2520\times 20=50400$

Explanation: We need to select 5 men from 7 men and 2 women from 3 women.

Number of ways to do this = ⁷C₅ × ³C₂ = ⁷C₂ × ³C₁ [∵ ⁿC_r = ⁿC_(n-r)]

$=63$

Explanation: 



The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and these 4 vowels must always come together. Hence these 4 vowels can be grouped and considered as a single letter. That is, MTHMTCS(AEAI).



Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times, 'T' occurs 2 times but rest of the letters are different.



Hence,number of ways to arrange these letters

$=\frac{8!}{(2!)(2!)}$



 $=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(2\times 1)(2\times 1)}=10080$



In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.



Number of ways to arrange these vowels among themselves $=\frac{4!}{2!}=\frac{4\times 3\times 2\times 1}{2\times 1}=12$



Hence, required number of ways

$=10080\times 12=120960$



Explanation: 



The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and these 4 vowels must always come together. Hence these 4 vowels can be grouped and considered as a single letter. That is, MTHMTCS(AEAI).



Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times, 'T' occurs 2 times but rest of the letters are different.



Hence,number of ways to arrange these letters

$=\frac{8!}{(2!)(2!)}$



 $=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(2\times 1)(2\times 1)}=10080$



In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.



Number of ways to arrange these vowels among themselves $=\frac{4!}{2!}=\frac{4\times 3\times 2\times 1}{2\times 1}=12$



Hence, required number of ways

$=10080\times 12=120960$

Explanation: 



We need to select 5 men from 8 men and 6 women from 10 women



Number of ways to do this

= 8C5 × 10C6

= 8C3 × 10C4 [∵ nCr = nC(n-r)]

$=\left(\frac{8\times 7\times 6}{3\times 2\times 1}\right)\left(\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\right)=56\times 210=11760$

Explanations



Explanation: 



The word 'LOGARITHMS' has 10 different letters.



Hence, the number of 3-letter words(with or without meaning) formed by using these letters

= 10P3

$=10\times 9\times 8=720$

Explanation: 



When a coin is tossed once, there are two possible outcomes: Head(H) and Tale(T)



Hence, when a coin is tossed 3 times, the number of possible outcomes

$=2\times 2\times 2=8$



(The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT )