Permutations and Combinations Test 34

Result

Permutations and Combinations Test 34

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Weekly Quiz Competition

Question 1
1 / -0

A committee of $$4$$ persons is to be formed from $$2$$ ladies, $$2$$ old men and $$4$$ young men such that it includes at least $$1$$ lady. at least $$1$$ old man and at most $$2$$ young men. Then the total number of ways in which this committee can be formed is :

A
$$40$$

B
$$41$$

C
$$16$$

D
$$32$$

Solution
Question 2
1 / -0

The number of values of 'r' satisfying the equation, $${^{39}C_{3r-1}}-{^{39}C_{r^2}}={^{39}C_{r^2-1}}-{^{39}C_{3r}}$$ is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 3
1 / -0

The number lock of a suitcase has four wheels, each labelled with 10-digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase

A
$$\dfrac { 1 }{ 5040 } $$

B
$$\dfrac { 3 }{ 5040 } $$

C
$$\dfrac { 7 }{ 5040 } $$

D
None of these

Solution
Question 4
1 / -0

An old man while dialing a $$7$$ digit telephone number remembers that the first four digits consists of one $$1's$$, one $$2's$$ and two $$3's$$. He also remembers that the fifth digits is either a $$4$$ or $$5$$ while has no memorising of the sixth digit, he remembers that the seventh digit is $$9$$ minus the sixth digit. Maximum number of distinct trials he has to try to make sure that he dials the correct telephone number, is

A
$$360$$

B
$$240$$

C
$$216$$

D
None of these

Solution

The telephone number has $$7$$ digits.
The first $$4$$ digits have one $$1$$, one $$2$$ and $$2$$ $$3$$'s.
$$\therefore$$ No. of ways of arranging the first $$4$$ digits$$=\dfrac{4!}{2!}=12$$ ways
The $$5^{th}$$ digit is a $$4$$ or a $$5$$.
$$\therefore$$ There are $$2$$ options for the $$5^{th}$$ digit.
Let the $$6^{th}$$ digit be x. Then, the $$7^{th}$$ digit is $$9-x$$.
$$\therefore$$ Possible combinations for $$6^{th}$$ and $$7^{th}$$ digits are:
($$6^{th}$$ digit, $$7^{th}$$ digit): $$(0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)$$.
$$\therefore$$ There are $$10$$ possible combinations for the $$6^{th}$$ and $$7^{th}$$ digits.
$$\therefore$$ No. of possible telephone numbers$$=12\times 2\times 10=240$$, to dial correct number.
$$\therefore$$ The correct answer is option (B).
Question 5
1 / -0

The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is-

A
672

B
640

C
512

D
None of these

Solution

choose any two of the seven digit
This may be done in $$^{7}C_{2}$$ ways
put 2 in these two digits .
The remaining 5 digit may be arranged
using 1 and 3 in $$2^{5}$$ ways.

so, squired numbers of numbers
=$$^{7}C_{2} \times 2^{5}$$
=$$\dfrac{7!}{5!2!}\times 2^{5}$$

=$$\dfrac{7\times 6\times 5!}{5!\times 2}\times 2^{5}$$

=$$21\times 32=672$$
Question 6
1 / -0

If $$^{8}C_{r}=^{8}C_{3}$$, then $$r$$ is equal to

A
$$5$$

B
$$4$$

C
$$8$$

D
$$6$$

Solution

$$^8C_r=^8C_3$$
$$\frac {8!}{r!(8-r)!}=\frac{8!}{3!\,5!}$$
$$\Rightarrow r!(8-r)!=3 !5!$$
$$\Rightarrow either \, r=3 \,\,\, or r=5 \Rightarrow (A)$$
Question 7
1 / -0

Find $$x$$, if $$\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}$$.

A
$$5$$

B
$$4$$

C
$$30$$

D
$$None$$

Solution
Question 8
1 / -0

If $$(1 + x)^n = \displaystyle \sum^{n}_{r = 0} {^nC_r} x^r$$ then $$C^2_0 + \dfrac{C^2_1}{2} + \dfrac{C^2_2}{3} + ... + \dfrac{C^2_n}{n + 1} =$$

A
$$\dfrac{{2n}!}{n !)^2}$$

B
$$\dfrac{{2n + 1}!}{{(n + 1)^2}!}$$

C
$$\dfrac{{2n - 1}!}{{(n + 1)^2}!}$$

D
$$\dfrac{{n}!}{{(n - 1)^2}!}$$
Question 9
1 / -0

When $$n!+1$$ is divided by any natural number between $$2$$ and $$n$$ then remainder obtained is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$
Question 10
1 / -0

The value of $$\sum _ { r = 1 } ^ { 5 } r \dfrac { ^ { n } C _ { r } } { ^ { n } C _ { r - 1 } } =?$$

A
$$5 ( n - 3 )$$

B
$$5 ( n - 2 )$$

C
$$5 \mathrm { n }$$

D
$$5 ( 2 n - 9 )$$

Solution