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Q:

# How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

 A) 216 B) 45360 C) 1260 D) 43200

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

$\inline \therefore$ The number of ways in which 9 letters can be arranged = $\inline \frac{9!}{2!\times 2!\times 2!}$ = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $\inline \frac{6!}{2!\times 2!}$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $\inline \frac{4!}{2!}$ = 12 ways.

$\inline \therefore$ The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Q:

Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is

 A) 1,00,000 B) 59,049 C) 3439 D) 6561

Explanation:

The numbers 0,1,2,3,4,5,6,7,8,9 are 10 in number while preparing telephone numbers any number can be used any number  of times.

$\inline \therefore$ This can be done in $\inline 10^{5}$ ways, but '0' is there

So, the numbers starting with '0' are to be excluded which one $\inline 9^{4}$ in number.

$\inline \therefore$ Total 5 digit telephone numbers = $\inline 10^{5}-9^{4}$ = 3439

1 421
Q:

In how many ways the letters of the word 'DESIGN' can be arranged so that no consonant appears at either of the two ends?

 A) 240 B) 72 C) 48 D) 36

Explanation:

DESIGN = 6 letters

No consonants appears at either of the two ends.

$\inline 2\times 4P_{4}$

=  2 x 4 x 3 x 2 x 1

=  48

4 404
Q:

In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together?

 A) 360 B) 720 C) 120 D) 840

Explanation:

CAPITAL = 7

Vowels = 3 (A, I, A)

Consonants = (C, P, T, L)

$\inline \therefore$5 letters which can be arranged in $\inline 5P_{5}=5!$

Vowels A,I = $\inline \frac{3!}{2!}$

$\inline \therefore$No.of arrangements = $\inline 5!\times \frac{3!}{2!}$=360

0 309
Q:

The number of ways that 8 beads of different colours be strung as a necklace is

 A) 2520 B) 2880 C) 4320 D) 5040

Explanation:

The number of ways of arranging n beads in a necklace is $\inline \frac{(n-1)!}{2}=\frac{(8-1)!}{2}=\frac{7!}{2}=2520$ (since n = 8)

5 547
Q:

The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is

 A) 7! x 7! B) 7! x 6! C) 6! x 6! D) 7! x 5!

The students should sit in between two teachers. There are 7 gaps in betweeen teachers when they sit in a round table. This can be done in $\inline 7P_{6}$ ways. 7 teachers can sit in (7-1)! ways.
$\inline \therefore$ Required no.of ways in $\inline 7P_{6}.6!$ = $\inline 7!.6!$