How many different words can be formed of the letter of the word combine so that vowels always remain together no two vowels are together?

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Answer

Long Answer

Solution not provided.Ans. (i) 720 (ii) 1440

(iii) 576

COMBINE HAS 7 LETTERS in which C,M,B,N R 4 CONSONANTS ND O,I,E R 3 VOWELS

1) OIE, C,M,B,N CAN BE ARRANGED 5!=120 ways

bt vowels can be arranged among themselves=3!=6 ways

total no.of ways=120*6=720

COMBINE HAS 7 LETTERS in which C,M,B,N R 4 CONSONANTS ND O,I,E R 3 VOWELS

1) OIE, C,M,B,N CAN BE ARRANGED 5!=120 ways

bt vowels can be arranged among themselves=3!=6 ways

total no.of ways=120*6=720

Answer

Verified

Hint: The word daughter has $8$ letters in which $3$ are vowels. For the vowels to always come together consider all the $3$ vowels to be one letter (suppose V) then total letters become $6$ which can be arranged in $6!$ ways and the vowels themselves in $3!$ ways.

Complete step-by-step answer:

Given word ‘DAUGHTER’ has $8$ letters in which $3$ are vowels and 5 are consonants. A, U, E are vowels and D, G, H, T, R are consonants.(i)We have to find the total number of words formed when the vowels always come together.Consider the three vowels A, U, E to be one letter V then total letters are D, G, H, T, R and V. So the number of letters becomes $6$So we can arrange these $6$ letters in $6!$ ways. Since the letter V consists of three vowels, the vowels themselves can interchange with themselves. So the number of ways the $3$vowels can be arranged is $3!$ Then,$ \Rightarrow $ The total number of words formed will be=number of ways the $6$ letters can be arranged ×number of ways the $3$ vowels can be arrangedOn putting the given values we get,$ \Rightarrow $ The total number of words formed=$6! \times 3!$ We know $n! = n \times \left( {n - 1} \right)! \times ...3,2,1$ $ \Rightarrow $ The total number of words formed=$6 \times 4 \times 5 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1$ On multiplying all the numbers we get, $ \Rightarrow $ The total number of words formed=$24 \times 5 \times 6 \times 6$ $ \Rightarrow $ The total number of words formed=$120 \times 36$ $ \Rightarrow $ The total number of words formed=$4320$

The number of words formed from ‘DAUGHTER’ such that all vowels are together is $4320$.

(ii)We have to find the number of words formed when no vowels are together.Consider the following arrangement- _D_H_G_T_RThe spaces before the consonants are for the vowels so that no vowels come together. Since there are $5$ consonants so they can be arranged in $5!$ ways.There are $6$ spaces given for $3$ vowels. We know to select r things out of n things we write use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$ So to select $3$ spaces of out $6$ spaces =${}^6{{\text{C}}_3}$ And the three vowels can be arranged in these three spaces in $3!$ ways.$ \Rightarrow $ The total number of words formed=${}^6{{\text{C}}_3} \times 3! \times 5!$$ \Rightarrow $ The total number of words formed=$\dfrac{{6!}}{{3!6 - 3!}} \times 5! \times 3!$ $ \Rightarrow $ The total number of words formed=$\dfrac{{6!}}{{3!}} \times 5!$ On simplifying we get-$ \Rightarrow $ The total number of words formed=$\dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} \times 5!$ $ \Rightarrow $ The total number of words formed=$120 \times 5 \times 4 \times 3 \times 2 \times 1$ On multiplying we get,$ \Rightarrow $ The total number of words formed=$14400$

The total number of words formed from ‘DAUGHTER’ such that no vowels are together is $14400$.

Note: Combination is used when things are to be arranged but not necessarily in order. Permutation is a little different. In permutation, order is important. Permutation is given by-

$ \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ Where n=total number of things and r=no. of things to be selected.

Both of your answers are incorrect.

Method 1: We arrange the consonants, then place the vowels in the spaces between and at the ends of the row.

There are $4!$ ways of arranging the four distinct consonants B, G, L, R. For each such arrangement, there are five spaces in which we could place the vowels, three between successive consonants and two at the ends of the row. To separate the vowels, we must place the three vowels in three of these five spaces. Choose two of the five spaces for the As and one of the remaining three spaces for the E. Thus, the number of admissible arrangements is $$4!\binom{5}{2}\binom{3}{1} = 720$$

Method 2: We use the Inclusion-Exclusion Principle.

There are seven letters in total, including two As, one B, one E, one G, one L, and one R. There are $\binom{7}{2}$ ways to choose two of the seven positions in the arrangement for the As. The remaining five letters can be placed in the remaining five positions in $5!$ ways. Hence, there are a total of $$\binom{7}{2}5! = \frac{7!}{2!5!} \cdot 5! = \frac{7!}{2!}$$ ways of arranging the letters of the word ALGEBRA.

From these, we must subtract those arrangements in which one or more pairs of vowels are consecutive.

A pair of vowels is consecutive: Two As are consecutive or an A and E are consecutive.

Two As are consecutive: We have six objects to arrange. They are AA, B, E, G, L, R. Since they are distinct, they can be arranged in $6!$ ways.

An A and an E are consecutive: We have six objects to arrange. They are B, E, G, L, R, and a block containing an A and an E. The six objects are distinct, so they can be arranged in $6!$ ways. The A and E can be arranged within the block in $2!$ ways. Hence, there are $6!2!$ such arrangements.

If we subtract the number of arrangements in which a pair of vowels is consecutive from the total, we will have subtracted each arrangement in which two pairs of vowels are consecutive twice, once for each way we could have designated one of those pairs as the pair of consecutive vowels. Since we only want to subtract such arrangements once, we must add them to the total.

Two pairs of consecutive vowels: Since there are only three vowels, this can only occur if the three vowels are consecutive. We have five objects to arrange, B, G, L, R, and the block of three vowels. Since the objects are distinct, they can be arranged in $5!$ ways. The three vowels can be arranged in three ways: AAE, AEA, EAA. Hence, the number of such arrangements is $5! \cdot 3$.

By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\frac{7!}{2!} - 6! - 6!2! + 5! \cdot 3 = 720$$