Question

The number of arrangements of the letters of the word ${}^{\prime }VEERBHADU{R}^{\prime }$in which two vowels are togerther and other two vowels are together but separated from each other-

• A. $15\times 6!$
• B. $18\times 7!$
• C. $21\times 6!$
• D. $21\times 7!$

Answer 1

Answer: (B)

$18\times 7!$

VEER BHADUR

(EE) (AU) and

(EU) (AE) are the only pair of Vowels possible.

VRBHDR can be arranged

in $\frac{6!}{2}$ ways (R is repeated)-------(1)

The vowel pairs can be placed in between the

spaces of the alphabets from (1)

For (EE) (AU) $\to$ 2 arrangements

(EU) (AE) $\to$ 4 arrangements

Number of arrangements

$=\frac{6!}{2}[7p_2\times 2+7p_2\times 4]$

$=\frac{6!}{2}[84+168]$

$=\frac{7!}{2}[12+24]=\frac{36\times 7!}{2}$

$=18\times 7!$ ways.