GO Classes Test Series 2024 | Discrete Mathematics | Test 4 | Question: 13

Answer 1

The possible positions for the two vowels are $(2, 4), (2, 5)$ and $(3, 5).$ Each of these results in two isolated consonants and two adjacent consonants. Thus the answer is the product of the following factors:

1. choose the vowel locations $(3\;\text{ways});$
2. choose the vowels $(2 \times 2 = 4 \;\text{ways});$
3. choose the isolated consonants $(3 \times 3 = 9 \;\text{ways}); $
4. choose the adjacent consonants $(3 \times 2 = 6\;\text{ways)}.$

The answer is $3 \times 4 \times 9 \times 6 = 648.$

This construction can be interpreted as a Cartesian product as follows. $\text{C1}$ is the set of lists of possible positions for the vowels, $\text{C2}$ is the set of lists of vowels in those positions, and $\text{C3}$ and $\text{C4}$ are sets of lists of consonants. Thus

• $\text{C1} = \{(2, 4),(2, 5),(3, 5)\}$
• $\text{C2} = \{\text{AA, AI, IA, II}\}$
• $C3 = \{\text{LL, LS, LT, SL, SS, ST, TL, TS, TT}\}$
• $C4 = \{\text{LS, LT, SL, ST, TL, TS}\}.$

For example, $((2,5), \text{IA, SS, ST})$ in the Cartesian product corresponds to the name $\text{SISTAS}.$

We can also solve this using the Rule of Sums. The possible vowel $\text{(V)}$ and consonant $\text{(C)}$ patterns for names are $\text{CCVCVC, CVCCVC}$ and $\text{CVCVCC.}$ Since these patterns are disjoint and cover all cases, we may compute the number of names of each type and add the results together. For the first pattern, we have a product of six factors, one for each choice of a letter$: 3\times 2 \times 2 \times 3 \times 2 \times 3 = 216.$ The other two patterns also give $216,$ for a total of $216 + 216 + 216 = 648$ names.

Comments

it also includes ((3,5), IA , SS ,ST) right then it will form the name SSITAS which is invalid , since SS are coming together. then the whole method is wrong.

please reply.

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@Student sv

When position 3,5 are Vowels, we have $2 \times 2$ ways for Vowels.

Position 4, 6 are consonants, hence, $3 \times 3$ ways for those positions.

Position 1,2 are consonants, hence, $3 \times 2$ ways for those positions (as they must be different).

So, answer is correct.

With ((3,5)IA , SS ,ST), we get the word STISAS, Not SSITAS.

Answer 2

CONDITIONS MENTIONED TO  DEFINE A NAME :- 

1. The name must be 6 alphabet long.
2. It must begin and end with a consonant.
3. It should contain 2 nonadjacent vowels.
4. Adjacent consonants must be different.

As per the above conditions the skeleton structure for the word is : C (p) (q) (r) (s) C , where p,q,r,s represent the position for  alphabets.

Now our aim is how can we populate these p,q,r as per the given conditions and for that we have only 3 cases.

1. CVCV – [ 2*2*3*2 ways ]  because for first consonant we only have 2 favorable consonants satisfying property 4.
2. VCVC – [2*3*2*2 ways ]
3. VCCV – [2*3*2*2 ways ]

// $2^{3}$*3 = 24

So in all 3 * [ 24 + 24 + 24 ] * 3 = 648 names possible.

 

Comments

Sorry to say, but you have left writing this constraint in your solution. 

"5th condition : Vowels are not adjacent."

 

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@Udhay_Brahmi u can edit his answer rather than writing here 😁

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I don't think I have powers to do so. 😅

But yeah, I think he got the point. 👍🏻