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From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if:

  1. Any letter can be repeated.
  2. No letter can be repeated.

My answer for the 1st part was $7 \cdot 7 \cdot 7 \cdot 7 \cdot 5 \cdot 5 \cdot 5$ (or $7^4 \cdot 5^3$). The way I thought about it is there are $4$ available slots for $7$ consonants that can be repeated so despite taking a consonant which we call it $c_1$, the second/third/fourth slot might also be $c_1$ so each of the $4$ slots has $7$ consonants to choose from so its $7 \cdot 7 \cdot 7 \cdot 7$. Same case for vowels which is $5 \cdot 5 \cdot 5$. Multiply those together to get total number of ways where the letter is repeated in a word which is $7^4 \cdot 5^3$.

However it turned out that the correct answer is actually $7^7 (7C4 \cdot 5C3)$. I don't understand the logic behind this. I don't get why not only the combinations nCr is used which is supposed to be used for the cases that don't require order and repetition, but also it is multiplied by $7^7$.

For the 2nd part, since the letter cannot be repeated then the available consonants and vowels decreases every time you choose from them so its $7 \cdot 6 \cdot 5 \cdot 4 \cdot 5 \cdot 4 \cdot 3$ (or $7P4 \cdot 5P3$).

However that's not the case because the correct answer is $7! \cdot (7C4 \cdot 5C3)$.

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    $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ Jul 3 at 13:36
  • $\begingroup$ @N.F.Taussig Thank you. Much appreciated. $\endgroup$
    – Farouq
    Jul 3 at 13:51

2 Answers 2

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From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if any letter can be repeated?

Your answer $7^4 \cdot 5^3$ is the number of ways the first four positions can be filled with consonants and the last three positions can be filled with vowels when letters may be repeated. However, the problem does not specify the positions of the consonants and vowels, so we must select them.

Notice that selecting which four of the seven positions will be filled with consonants also determines the positions of the three vowels since each of the remaining three positions must be filled with vowels.

We must select which four of the seven positions must be filled with consonants, which can be done in $\binom{7}{4}$ ways. Each of those four selected positions can be filled with a consonant in seven ways. Each of the remaining three positions can be filled with a vowel in five ways. Hence, there are $$\binom{7}{4}7^4 \cdot 5^3$$ words that can be formed consisting of four consonants and three vowels selected from an alphabet of seven consonants and three vowels with repetition.

The stated answer $7^7\dbinom{7}{4}\dbinom{5}{3}$ is nonsense.

From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if no letter can be repeated?

Your answer $P(7, 4)P(5, 3)$ is the number of ways of filling the first four positions with distinct consonants and the last three positions with distinct vowels. However, the problem does not specify the positions of the consonants and vowels, so we must select them.

As above, we select four of the seven positions for the consonants. There are $P(7, 4)$ ways to arrange four of the seven consonants in those positions and $P(5, 3)$ ways to arrange three of the five vowels in the remaining three positions. Hence, there are $$\binom{7}{4}P(7, 4)P(5, 3)$$ words that can be formed consisting of four distinct consonants and three distinct vowels selected from an alphabet of seven consonants and five vowels.

Notice that $$\binom{7}{4}P(7, 4)P(5, 3) = \frac{7!}{4!3!} \cdot \frac{7!}{3!} \cdot \frac{5!}{2!} = 7! \cdot \frac{7!}{4!3!} \cdot \frac{5!}{2!3!} = 7!\binom{7}{4}\binom{5}{3}$$ The author of your book arrived at the answer by selecting four of the seven consonants in $\binom{7}{4}$ ways, selecting three of the five vowels in $\binom{5}{3}$ ways, and then arranging the seven distinct selected letters in $7!$ ways.

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  • $\begingroup$ I'm still quite confused. Why does this $7^{4} \cdot 5^{3}$ $only$ $means$ the first four positions with consonants and the last three positions with vowels? Isn't multiplication commutative? Since it's commutative, then the consonants and vowels should be positioned anywhere because $7^{4} \cdot 5^{3} = 7^{2} \cdot 5^{1} \cdot 7^{2} \cdot 5^{2} = 5^{3} \cdot 7^{4} $ etc... $\endgroup$
    – Farouq
    Jul 3 at 15:21
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    $\begingroup$ Let's consider a simpler problem. Suppose we wish to form three-letter words with two consonants and one vowel. The consonants can be in the first and second positions, first and third positions, or second and third positions, giving $7 \cdot 7 + 5 + 7 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 7 = 3 \cdot 7^2 \cdot 5 = \binom{3}{2}7^2 \cdot 5$ possible words. The factor of $\binom{3}{2}$ in this example is the number of ways of choosing positions for the consonants. Your method only counts the first term in that sum. $\endgroup$ Jul 3 at 16:51
  • $\begingroup$ I think the picture is much more clearer now. Thank you. $\endgroup$
    – Farouq
    Jul 3 at 18:16
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My result differs from yours and the first also from what you refer to as the correct answer.

1)

Start with $7$ open spots and choose $4$ of them to be filled up with a consonant. The others are to be filled up with vowels.

There are $7^4$ ways for the consonants and $5^3$ ways for the vowels.

That gives a total of $$\binom74\times7^4\times5^3$$

2.

If repetition is not allowed then there are $7\times6\times5\times4$ ways for the consonants and $5\times4\times3$ ways for the vowels.

That gives a total of $$\binom74\times(7\times6\times5\times4)\times(5\times4\times3)$$

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    $\begingroup$ Your second answer agrees with the stated answer. The stated answer for the first question is nonsense. $\endgroup$ Jul 3 at 14:15
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    $\begingroup$ @N.F.Taussig I see now. Thank you for attending me. I was too lazy to check properly myself :-). $\endgroup$
    – drhab
    Jul 3 at 14:19

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