Solution: (e)

C R E A M

1  2 3  4  5

Required number of ways = \(\displaystyle 5!=5\times 4\times 3\times 2\times 1=120\)

Solution: (d)

No. of vowels in the word THERAPY = 2 i.e. E and A

In such cases we treat the group of two vowels as one entity or one letter because they are supposed to always come together. Thus, the problem reduces to arranging 6 letters i.e. T, H, R, P, Y and EA in 6 vacant places.

No. of ways 6 letters can be arranged in 6 places = \(\displaystyle 6!\)

\(\displaystyle 6\times 5\times 4\times 3\times 2\times 1=720\)

But the vowels can be arranged themselves in 2 different ways by interchanging their position. Hence, each of the above 720 arrangements can be written in 2 ways.

Therefore Required no. of total arrangements when two vowels are together

= \(\displaystyle 720\times 2=1440\)

Total no. of arrangements of THERAPY = \(\displaystyle 7!\)

\(\displaystyle 7\times 6\times 5\times 4\times 3\times 2\times 1=5040\)

No. of arrangement when vowels do not come together

\(\displaystyle 5040-1440=3600\)

Alternate direct method

Given word is THERAPY.

Number of letters in the given word = 7

These 7 letters can be arranged in 7! ways.

Number of vowels in the given word = 2 (E, A)

The number of ways of arrangement in which vowels come together is 6! x 2! ways

Hence, the required number of ways can the letters of the word ‘THERAPY’ be arranged so that the vowels never come together = 7! – (6! x 2!) ways = 5040 – 1440 = 3600 ways.

Solution: (a)

The word SACRED consists of 4 consonants (SCRD) and two vowels (AE). On keeping vowels together we get SCRD (AE).

Therefore, Number of arrangements = \(\displaystyle 5!\times 2!=5\times 4\times 3\times 2\times 1\times 1\times 2=240\)

Solution: (c)

The word ‘LEADING’ has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, \(\displaystyle 5(4+1=5)\) letters can be arranged in \(\displaystyle 5!=120\) ways.

The vowels (EAI) can be arranged among themselves in \(\displaystyle 3!=6\) ways.

Therefore, Required number of ways = \(\displaystyle (120\times 6)=720\)

Solution: (a)

The word RUMOUR consists of 6 letters in which each of R and U comes twice.

Therefore, Number of arrangements = \(\displaystyle \frac{{6!}}{{2!2!}}\)

\(\displaystyle \frac{{6\times 5\times 4\times 3\times 2\times 1}}{{2\times 2}}=180\)