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# What is the total no of ways of selecting atleast one item from each of the two sets containing 6 different items each?

 A) 2856 B) 3969 C) 480 D) None of these

Explanation:

We can select atleast one item from 6 different items = $\inline&space;(2^{6}-1)$

Similarly we can select atleast one item from other set of 6 different items in $\inline&space;(2^{6}-1)$ ways.

$\therefore$ Required number of ways = $\inline&space;(2^{6}-1)(2^{6}-1)$

= $\inline&space;(2^{6}-1)^2$ = 3969

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• Related Questions

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

Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

1

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

Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

4

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

Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

0

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)

Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

5

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 $\inline 7P_{6}.6!$ = $\inline 7!.6!$