Permutation Test

Result

Permutation Test - 7

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
1 / -0

If ^{m + n}P₂ = 90 and ^{m – n}P₂ = 30, then (m, n) is given by -

A
(7, 3)

B
(16, 8)

C
(9, 2)

D
(8, 2)

Solution

⇒ (m + n) (m + n – 1) = 90 = 10 × 9
⇒ m + n = 10 ……(1)
and (m – n) (m – n – 1) = 30 = 6 × 5
⇒ m – n = 6 ……(2)
solving (1) & (2) we get
m = 8, n = 2
Question 2
1 / -0

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Determine the number of words which have at least one letter repeated.

A
69762

B
69676

C
69760

D
69766

Solution

All 5 letters word is = 100000
All five letters word which have no any letter repeated is = 10 × 9 × 8 × 7 × 6 = 30240
Number of words which have at least one letter repeated is
= 100000 – 30240 = 69760
Question 3
1 / -0

How many numbers of four digits greater than 2300 can be formed with the digits 0, 1, 2, 3, 4, 5 and 6; no digit being repeated in any number?

A
560

B
590

C
90

D
360

Solution

When number started greater then 2 then
Total number = 4 × 6 × 5 × 4 = 480
when number started from 2 then
numbers = 1 × 4 × 5 × 4 = 80
Total number = 480 + 80 = 560
Question 4
1 / -0

There are m men and n monkeys (n > m). If a man have any number of monkeys. In how many ways may every monkey have a master?

A
n^m

B
mⁿ

C
mⁿ – n^m

D
mn

Solution

Total ways
= m × m × m ……. × m = mⁿ
n times
Question 5
1 / -0

How many 7 digit numbers can be written using three digits 1, 2 and 3 under the condition that the digits 2 occurs twice in each number ?

A
⁷P₂ × 2⁵

B
⁷C₂ × 2⁵

C
⁷P₂× 2³

D
⁷C₂ × 2³

Solution

7 places can be chosen by fixing digit 2 at two places (There are two 2, which cannot be arranged since both are like)
= ⁷C₅
= ⁷C₂
But two remaining digits 1 and 3 can occupy 5 places in 2⁵ ways.
∴ Required no. of ways = ⁷C₂ × 2⁵