Permutations and Combinations Test 32

Result

Permutations and Combinations Test 32

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Question 1
1 / -0

$$\sum _{ k=1 }^{ 10 }{ k.k! } =$$

A
10!

B
11!

C
10!+1

D
11!-1

Solution

Given,

$$\sum _{k=1}^{10}kk!$$

$$a_k=kk!$$

$$a_1=1\cdot \:1!=1$$

$$a_2=2\cdot \:2!=4$$

Similarly,

$$a_3=18$$

$$a_4=96$$

$$a_5=600$$

$$a_6=4320$$

$$a_7=35280$$

$$a_8=322560$$

$$a_9=3265920$$

$$a_{10}=36288000$$

$$=1+4+18+96+600+4320+35280+322560+3265920+36288000$$

$$=39916799$$

$$=11!-1$$
Question 2
1 / -0

Solve:
$$\dfrac{^{n}C_{r}}{^{n}C_{r-1}}=$$

A
$$\dfrac{n-r}{r}$$

B
$$\dfrac{n+r-1}{r}$$

C
$$\dfrac{n-r+1}{r}$$

D
$$\dfrac{n-r-1}{r}$$

Solution
Question 3
1 / -0

The number of was dividing $$52$$ cards amongst four players equally, are

A
$$\dfrac {52!}{(13!)^{4}}$$

B
$$\dfrac {52!}{(13!)^{4}4!}$$

C
$$\dfrac {52!}{(12!)^{4}4!}$$

D
None of these
Question 4
1 / -0

A double Decker is can accommodate 20 passengers 7 in the lower deck 13 in the upper deck. The number of ways the passengers can be accommodate if 5 want to sit only in lower deck and 8 want to sit only in upper deck is

A
$$^{7}C_{5}$$

B
$$^{7}C_{3}$$

C
$$^{7}C_{1}$$

D
$$^{7}C_{6}$$

Solution
Question 5
1 / -0

The number of words which can be made out of the letters of the word $$'MOBILE'$$ when consonants always occupy odd places is _______.

A
$$20$$

B
$$36$$

C
$$30$$

D
$$720$$

Solution
Question 6
1 / -0

If $$^{n}C_{4},\ ^{n}C_{5}$$ and $$^{n}C_{6}$$ in A.P., then possible value of $$n$$ is

A
$$6$$

B
$$12$$

C
$$14$$

D
$$21$$

Solution
Question 7
1 / -0

How many 10 digits number can be written by using digits (9 and 2) ?

A
$$^{10}C_1 + ^9C_2$$

B
$$2^{10}$$

C
$$^{10} C_2$$

D
$$10 !$$

Solution

Total number of numbers =2×2.....×10 times
⇒$$2^{10}$$
Hence (B) is the correct answer.
Question 8
1 / -0

If $$CARPET$$ is coded as $$TCEAPR$$ then the code for $$NATIONAL$$ would be written as

A
$$NLATNOIA$$

B
$$LANOITAN$$

C
$$LNAANTOI$$

D
$$LNOINTAA$$

Solution
Question 9
1 / -0

All possible three digits even numbers which can be formed with the condition that if $$5$$ is one of the digit, then $$7$$ is the next digit is:

A
$$5$$

B
$$325$$

C
$$345$$

D
$$365$$

Solution

An even number can be formed by using one of the digits $$0, 2, 4, 6$$ or $$8$$ at the units place.
$$5$$ can not be at ten's place, since in that case $$7$$ has to be placed at the units place. In that case, the number formed is not an even number.
When $$5$$ is at the hundred's place, then $$7$$ will be at the ten's place. The units place can be filled in 5 ways by using one of the digits $$0, 2, 4, 6$$ or $$8.$$
$$\Rightarrow$$ Number of even numbers in this case $$= 1 \times 1 \times 5 = 5$$
When $$5$$ is not at the hundred's place, so the hundred's place can be filled in $$8$$ ways ($$0$$ and $$5$$ cannot be used at hundred's place).
Ten's place can be filled in $$9$$ ways ($$5$$ cannot be used at ten's place).Units place can be filled in $$5$$ ways ($$0, 2, 4, 6,$$ or $$8$$ can occupy the even place).
$$\Rightarrow$$ Number of even number in this case $$= 8 \times 9 \times 5 = 360$$
Hence, by fundamental principle of addition,
$$\Rightarrow$$ Total number of even number $$= 360 + 5 = 365.$$
Question 10
1 / -0

A committee of $$10$$ is to be formed from $$8$$ women and $$6$$ men. In how many of these committees the women are in majority?

A
$$515$$

B
$$545$$

C
$$575$$

D
$$595$$

Solution

for majority, women should be $$ > 5$$ in no.
$$\therefore $$ selecting the committee $$\Rightarrow $$
$$^{8}C_{6}\times ^{6}C_{4}+^{8}C_{7}\times ^{6}C_{3}+^{8}C_{8}\times ^{6}C_{2}$$
$$=\frac{8\times 7\times 6\times 5}{2\times 1\times 2\times 1}+\frac{8}{1}\times \frac{6\times 5\times 4}{3\times 2\times 1}+\frac{8}{8}\times \frac{6\times 5}{2\times 1}$$
$$= 420+160+15$$
$$=595$$