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A Learning Portal from Recruitment India

A.

252

B.

250

C.

1001

D.

Answer with explanation

Answer: Option CExplanation

The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it.

This case can be represented as arranging ten balls and (5 – 1) four walls in the single row, which can be done in ^{14}C_{4} ways. (The balls placed between every successive pair of walls belong to one group) ^{14}C_{4} = 1001 ways.

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In a cricket championship, there are 21 matches. If each team plays one match with every other team, the number of teams is

A.

7

B.

9

C.

10

D.

None of these

Answer with explanation

Answer: Option AExplanation

Let n be the number of teams.

^{n}C_{2} = 21

(n(n-1)/2) = 21

⇒ n(n-1) = 42 ∴

⇒ n = 7

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If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

A.

601

B.

600

C.

603

D.

602

Answer with explanation

Answer: Option AExplanation

If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

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How many numbers greater than a million can be formed by using the digits 7, 4, 6 and 0 if 4 has to be used twice, 6 has to be used thrice and the rest only once?

A.

B.

720

C.

360

D.

120

Answer with explanation

Answer: Option CExplanation

The given digits are 4, 4, 6, 6, 6, 7 and 0.

Totally we have 7 digits. So numbers greater than a million can be formed by using all the digits.

4 occurs twice, 6 occurs thrice while 0 and 7 once,

Therefore total number of arrangements = 7! / (3! x 2!) = 420 We have to avoid 0 in the starting place,

Number of ways in which 0 can be in the first place is 6! / 2! X 3! = 60

Hence the total number of ways = 420 – 60 = 360

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The sum of the divisors of 2^{6}.3^{4}.5^{7} is

A.

2^{6}.3^{8}.5^{4}.7^{3}

B.

2^{6}.3^{8}.5^{4}.7^{3}-1

C.

2^{7}.3^{5}.5^{8}-2.3.5

D.

None of these

Answer with explanation

Answer: Option DExplanation

Any divisor of 2^{6}.3^{4}.5^{7}. is of the form 2^{a}.3^{b}.5^{c}. where 0≤ a ≤ 6, 0 ≤ b ≤ 4 & 0 ≤ c ≤ 7.

Thus the sum of the divisors of 2^{6}.3^{4}.5^{7}.

is (1 + 2+…..+2^{5} + 2^{6})(1 + 3 + …. + 3^{4})(1 + 5 +….+ 5^{7}) =

((2^{7}-1)/(2-1))((3^{5}-1)/(3-1))((5^{8}-1)/(5-1))

=((2^{7}-1)(3^{5}-1)(5^{8}-1)/2.4)

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A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior?

A.

²²C₁₀

B.

²²C₁₀ + 1

C.

²²C₉ + ¹⁰C₁

D.

²²C₁₀ – 1

Answer with explanation

Answer: Option DExplanation

The total number of ways of forming the group of ten representatives is ²²C₁₀.

The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way

The required number of ways = ²²C₁₀ – 1

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How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?

A.

40

B.

400

C.

5040

D.

2520

Answer with explanation

Answer: Option CExplanation

‘LOGARITHMS’ contains 10 different letters.

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A.

21530

B.

8! × 360

C.

8! × 480

D.

8! × 240

Answer with explanation

Answer: Option CExplanation

Total no. of seats,

= 1 grandfather + 5 sons and daughters + 8 grandchildren

= 14

The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways.

The grandfather can occupy a seat in (5 – 1) = 4 ways (4 gaps between 5 sons and daughter).

And, the remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter).

Hence total number of required ways,

= 8! × 480

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From a pack of 52 cards, 3 cards are drawn together at random, What is the probability of both the cards are a king?

A.

1/5225

B.

1/5525

C.

5525

D.

1/525

Answer with explanation

Answer: Option BExplanation

Let S be the sample space

n(S) = 52C_{3} = [( 52 x 51 x 50) / (3 x 2 x 1)] = 13600 / 6 = 22100

Let E = event of getting 3 kings out of 4.

n(E) = 4C_{3} = [(4 x 3 x 2 x 1) / (3 x 2 x 1)] = 24/6 = 4

therefore, p(E) = [n(E) / n(S)]

==> p(E) = [ 4 / 22100]

==> p(E) = 1/5525

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16 persons are participated in a party. In how many differentways can they host the seat in a circular table, if the 2particular persons are to be seated on either side of the host?

A.

16! × 2

B.

14! × 2

C.

18! × 2

D.

14!

Answer with explanation

Answer: Option BExplanation

(16 – 2)! × 2 = 14! × 2

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How many 3-digit numbers can be formed from the digit 2,3,5,6,7 and 9, which are divisible by 5 and none of the digits is repeated?

A.

5

B.

10

C.

15

D.

20

Answer with explanation

Answer: Option DExplanation

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

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