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Permutations and Combinations: FormulasPermutations and Combinations: Exercise - 1Permutations and Combinations: Exercise - 2

1How many arrangements can be made of the letters of the word “ASSASSINATION”? In how many of them are the vowels always together?A$\dfrac{{13!}}{{{{\left( {4!} \right)}^2}}},\;\dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}}$

B$\dfrac{{13!}}{{6! \times 7!}},\;\dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}}$

C$\dfrac{{13!}}{{6! \times 7!}},\;\dfrac{{8! \times 6!}}{{6! \times 7!}}$

D$\dfrac{{13!}}{{{{\left( {4!} \right)}^2}}},\;\dfrac{{8! \times 6!}}{{6! \times 7!}}$

Answer: A

Explanation:

Total letters in the word $ASSASSINATION$ $=13$

$(SSSS), (AAA), (II), (NN), T, O$

Formula:Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and so on $ = \dfrac{{n!}}{{p! \times q! \times ...}}$

Total number of arrangements $ = \dfrac{{13!}}{{4! \times 3! \times 2! \times 2!}}$ $ = \dfrac{{13!}}{{4! \times 4 \times 3!}}$ $ = \dfrac{{13!}}{{{{\left( {4!} \right)}^2}}} \qquad (1)$

Vowels in the given word $=AAAIIO$

Let us put them in a box and name it a letter $X$.

$\boxed{AAAIIO}_{\rightarrow{X}}, S, S, S, S, N, N, T $

Now total letters $= 8$

Number of ways of arranging $8$ letters in $8$ places $ = \dfrac{{8!}}{{4! \times 2!}}$

Numebr of ways the vowels in the box arrange themselves $ = \dfrac{{6!}}{{3! \times 2!}}$

Total ways in which all vowels together $ = \dfrac{{8!}}{{4! \times 2!}} \times \dfrac{{6!}}{{3! \times 2!}}$ $ = \dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}} \qquad (2)$

Total letters in the word $ASSASSINATION$ $=13$

$(SSSS), (AAA), (II), (NN), T, O$

Formula:Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and so on $ = \dfrac{{n!}}{{p! \times q! \times ...}}$

Total number of arrangements $ = \dfrac{{13!}}{{4! \times 3! \times 2! \times 2!}}$ $ = \dfrac{{13!}}{{4! \times 4 \times 3!}}$ $ = \dfrac{{13!}}{{{{\left( {4!} \right)}^2}}} \qquad (1)$

Vowels in the given word $=AAAIIO$

Let us put them in a box and name it a letter $X$.

$\boxed{AAAIIO}_{\rightarrow{X}}, S, S, S, S, N, N, T $

Now total letters $= 8$

Number of ways of arranging $8$ letters in $8$ places $ = \dfrac{{8!}}{{4! \times 2!}}$

Numebr of ways the vowels in the box arrange themselves $ = \dfrac{{6!}}{{3! \times 2!}}$

Total ways in which all vowels together $ = \dfrac{{8!}}{{4! \times 2!}} \times \dfrac{{6!}}{{3! \times 2!}}$ $ = \dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}} \qquad (2)$

2In how many ways can the letters of the word ARRANGE be arranged so that two R’s are never together

A900

B360

C120

D1260

Answer: A

Explanation:

$ARRANGE$ $=(AA), (RR), N, G, E$

Two R’s are never together $=$ (Total possible arrangements) $-$ (Two R’s are always together)

Total number of arrangements = $\dfrac{{7!}}{{2! \times 2!}}$ = $\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}}$ $= 1260$

Arrangements with both R’s together:

Let us put both $R$'s are in a box and name it letter $X$.

$\boxed{RR}_{\rightarrow{X}}, A, A, N, G, E $

Number of ways of arranging above $6$ letters = $\dfrac{{{\text{6}}!}}{{{\text{2}}!}}$ $= 360$

(Note: Two $R$'s in the box arrange only one way)

R’s never together $= 1260 - 360 = 900$

$ARRANGE$ $=(AA), (RR), N, G, E$

Two R’s are never together $=$ (Total possible arrangements) $-$ (Two R’s are always together)

Total number of arrangements = $\dfrac{{7!}}{{2! \times 2!}}$ = $\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}}$ $= 1260$

Arrangements with both R’s together:

Let us put both $R$'s are in a box and name it letter $X$.

$\boxed{RR}_{\rightarrow{X}}, A, A, N, G, E $

Number of ways of arranging above $6$ letters = $\dfrac{{{\text{6}}!}}{{{\text{2}}!}}$ $= 360$

(Note: Two $R$'s in the box arrange only one way)

R’s never together $= 1260 - 360 = 900$

3In how many ways can the letters of the word ARRANGE be arranged so that The Two A’s are together but not two R’s

A900

B360

C240

D1260

Answer: C

Explanation:

The Two A's are together but not two R's ＝ Two A's are always together $-$ (Two R's and two A's are always together)

Arrangements with both A's together:

Let us put both $A$'s are in a box and name it letter $X$.

$\boxed{AA}_{\rightarrow{X}}, R, R, N, G, E $

Number of ways of arranging above $6$ letters ＝ $\dfrac{{{\text{6}}!}}{{{\text{2}}!}}$ ＝ $360$

(Note: Two $A$'s in the box arrange only one way)

Arrangements with both A's and R's together:

Let us put both $R$'s are in a box and name it letter $X$ and both $A$'s are in a box and name it letter $Y$

$\boxed{RR}_{\rightarrow{X}},\boxed{AA}_{\rightarrow{Y}}, N, G, E$

Number of ways of arranging above $5$ letters $=5!$ $= 120$

(Note: Two $A$'s and two $R$'s in the boxes arrange themselves only one way)

The Two A's are together but not two R's ＝ $360 - 120 = 240$

The Two A's are together but not two R's ＝ Two A's are always together $-$ (Two R's and two A's are always together)

Arrangements with both A's together:

Let us put both $A$'s are in a box and name it letter $X$.

$\boxed{AA}_{\rightarrow{X}}, R, R, N, G, E $

Number of ways of arranging above $6$ letters ＝ $\dfrac{{{\text{6}}!}}{{{\text{2}}!}}$ ＝ $360$

(Note: Two $A$'s in the box arrange only one way)

Arrangements with both A's and R's together:

Let us put both $R$'s are in a box and name it letter $X$ and both $A$'s are in a box and name it letter $Y$

$\boxed{RR}_{\rightarrow{X}},\boxed{AA}_{\rightarrow{Y}}, N, G, E$

Number of ways of arranging above $5$ letters $=5!$ $= 120$

(Note: Two $A$'s and two $R$'s in the boxes arrange themselves only one way)

The Two A's are together but not two R's ＝ $360 - 120 = 240$

4In how many ways can the letters of the word ARRANGE be arranged so that neither Two A’s nor two R's are together

A900

B360

C120

D660

Answer: D

Explanation:

Neither 2 A’s or 2R’s are together ＝ Total possible arrangements $-$ (two A’s are always together) $-$ (Two R's are always together) $+$ (Both Two A's and Two R's are always together)

Total number of arrangements ＝ $\dfrac{{7!}}{{2! \times 2!}}$ ＝ $\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}}$ ＝ $1260$

Number of arrangements of A's are together ＝ $360$ ($\because $ refer to the above two questions)

Number of arrangements of R's are together ＝ $360$ ($\because $ refer to the above two questions)

Number of arrangements of both A's are together and both R's are together ＝ $120$ ($\because $ refer to the above two questions)

Neither 2 A’s or 2R’s are together ＝ $1260 - 360 - 360 + 120 = 660$

Neither 2 A’s or 2R’s are together ＝ Total possible arrangements $-$ (two A’s are always together) $-$ (Two R's are always together) $+$ (Both Two A's and Two R's are always together)

Total number of arrangements ＝ $\dfrac{{7!}}{{2! \times 2!}}$ ＝ $\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}}$ ＝ $1260$

Number of arrangements of A's are together ＝ $360$ ($\because $ refer to the above two questions)

Number of arrangements of R's are together ＝ $360$ ($\because $ refer to the above two questions)

Number of arrangements of both A's are together and both R's are together ＝ $120$ ($\because $ refer to the above two questions)

Neither 2 A’s or 2R’s are together ＝ $1260 - 360 - 360 + 120 = 660$

5Ten different alphabets are given. Words containing five alphabets are to be formed from them. Find the number of words which have exactly one alphabet repeats.

A${}^{10}{P_5}$

B${\rm{10}}^{\rm{5}} $

C${10^5}{ - ^{10}}{P_5}$

D$58060$

Answer: D

Explanation:

We have 10 alphabets, we are to form five alphabet word, and given that repetition of exactly one alphabet is allowed. We can choose that alphabet in 10 ways. Now, following cases are possible:

1. It repeats exactly two times: $AABCD$ =

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of selecting remaining three alphabets $ = {}^9{C_3}$

Number of ways of arranging $AABCD$ letters $ = \dfrac{{5!}}{{2!}}$

Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_3} \times \dfrac{{5!}}{{2!}}$ $= 50400$

2. It repeats exactly three times:$AAABC$

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of selecting remaining three alphabets $ = {}^9{C_2}$

Number of ways of arranging $AAABC$ letters $ = \dfrac{{5!}}{{3!}}$

Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_2} \times \dfrac{{5!}}{{3!}}$ $= 7200$

3. It repeats exactly four times:$AAAAB$

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of selecting remaining three alphabets $ = {}^9{C_1}$

Number of ways of arranging $AAAAB$ letters $ = \dfrac{{5!}}{{4!}}$

Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_1} \times \dfrac{{5!}}{{4!}}$ $= 450$

4. It repeats exactly five times:$AAAAA$

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of arranging $AAAAA$ letters $ = 1$

Total numbers of ways $ = {}^{10}{C_1} \times 1$ $= 10$

Total number of words with only one alphabet repeating $= 58060$

Note:

Total number of words that can be formed if repetition is not allowed ＝ ${}^{10}{P_5}$

If repetition is allowed total number of words that can be formed will be ＝ ${\rm{10}}^{\rm{5}} $

The number of arrangements having atleast one alphabet repeating ＝ ${10^5}{ - ^{10}}{P_5}$

We have 10 alphabets, we are to form five alphabet word, and given that repetition of exactly one alphabet is allowed. We can choose that alphabet in 10 ways. Now, following cases are possible:

1. It repeats exactly two times: $AABCD$ =

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of selecting remaining three alphabets $ = {}^9{C_3}$

Number of ways of arranging $AABCD$ letters $ = \dfrac{{5!}}{{2!}}$

Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_3} \times \dfrac{{5!}}{{2!}}$ $= 50400$

2. It repeats exactly three times:$AAABC$

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of selecting remaining three alphabets $ = {}^9{C_2}$

Number of ways of arranging $AAABC$ letters $ = \dfrac{{5!}}{{3!}}$

Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_2} \times \dfrac{{5!}}{{3!}}$ $= 7200$

3. It repeats exactly four times:$AAAAB$

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of selecting remaining three alphabets $ = {}^9{C_1}$

Number of ways of arranging $AAAAB$ letters $ = \dfrac{{5!}}{{4!}}$

Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_1} \times \dfrac{{5!}}{{4!}}$ $= 450$

4. It repeats exactly five times:$AAAAA$

Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$

Number of ways of arranging $AAAAA$ letters $ = 1$

Total numbers of ways $ = {}^{10}{C_1} \times 1$ $= 10$

Total number of words with only one alphabet repeating $= 58060$

Note:

Total number of words that can be formed if repetition is not allowed ＝ ${}^{10}{P_5}$

If repetition is allowed total number of words that can be formed will be ＝ ${\rm{10}}^{\rm{5}} $

The number of arrangements having atleast one alphabet repeating ＝ ${10^5}{ - ^{10}}{P_5}$

6How many 4 letter words may be formed by using the letters of the word 'ASSASSINATION'

A916

B917

C360

D480

Answer: B

Explanation:

$(SSSS), (AAA), (II), (NN), T, O$

We have above 6 different letters.

We can choose 4 letter words in the following ways.

a. All the 4 letters are different - $ABCD$

Number of ways of choosing 4 letters from 6 different letters ＝ ${}^6{C_4}$

Number of ways of arranging $ABCD$ letters ＝ $4!$

Total ways ＝ ${}^6{C_4} \times 4!$

b. 2 letters same and 2 are different - $AABC$

Number of ways of choosing the repeated letter = ${}^4{C_1}$

Number of ways of choosing remaining two letters from 5 different letters ＝ ${}^5{C_2}$

Number of ways of arranging $ABCD$ letters ＝ $ = \dfrac{{4!}}{{2!}}$

Total ways ＝ ${}^4{C_1} \times {}^5{C_2} \times \dfrac{{4!}}{{2!}}$

c. 2 letters repeat twice - $AABB$

Number of ways of choosing the two repeated letters = ${}^4{C_2}$

Number of ways of arranging $AABB$ letters ＝ $ = \dfrac{{4!}}{{2! \times 2!}}$

Total ways ＝ ${}^4{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$

d. 1 letter repeat thrice - $AAAB$

Number of ways of choosing the repeated letter = ${}^2{C_1}$

Number of ways of choosing remaining letter from 5 different letters ＝ ${}^5{C_1}$

Number of ways of arranging $ABCD$ letters ＝ $ = \dfrac{{4!}}{{3!}}$

Total ways ＝ ${}^2{C_1} \times {}^5{C_1} \times \dfrac{{4!}}{{3!}}$

e. 1 letter repeat 4 times - $AAAA$

We choose the letter in only one way (only S we can choose) and we arrange them in 1 way.

Total ways $= 360 + 480 + 36 + 40 + 1 = 917$

$(SSSS), (AAA), (II), (NN), T, O$

We have above 6 different letters.

We can choose 4 letter words in the following ways.

a. All the 4 letters are different - $ABCD$

Number of ways of choosing 4 letters from 6 different letters ＝ ${}^6{C_4}$

Number of ways of arranging $ABCD$ letters ＝ $4!$

Total ways ＝ ${}^6{C_4} \times 4!$

b. 2 letters same and 2 are different - $AABC$

Number of ways of choosing the repeated letter = ${}^4{C_1}$

Number of ways of choosing remaining two letters from 5 different letters ＝ ${}^5{C_2}$

Number of ways of arranging $ABCD$ letters ＝ $ = \dfrac{{4!}}{{2!}}$

Total ways ＝ ${}^4{C_1} \times {}^5{C_2} \times \dfrac{{4!}}{{2!}}$

c. 2 letters repeat twice - $AABB$

Number of ways of choosing the two repeated letters = ${}^4{C_2}$

Number of ways of arranging $AABB$ letters ＝ $ = \dfrac{{4!}}{{2! \times 2!}}$

Total ways ＝ ${}^4{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$

d. 1 letter repeat thrice - $AAAB$

Number of ways of choosing the repeated letter = ${}^2{C_1}$

Number of ways of choosing remaining letter from 5 different letters ＝ ${}^5{C_1}$

Number of ways of arranging $ABCD$ letters ＝ $ = \dfrac{{4!}}{{3!}}$

Total ways ＝ ${}^2{C_1} \times {}^5{C_1} \times \dfrac{{4!}}{{3!}}$

e. 1 letter repeat 4 times - $AAAA$

We choose the letter in only one way (only S we can choose) and we arrange them in 1 way.

Total ways $= 360 + 480 + 36 + 40 + 1 = 917$

7How many numbers greater than a million can be formed by using the digits 2, 3, 0, 3, 5, 2, 3 which will be divisible by 5?

A110

B420

C720

D490

Answer: A

Explanation:

We have 7 digits: $(2, 2), (3, 3, 3), 0, 5, 2$

We have to form the numbers greater than 1,000,000.

$\quad \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}$

Numbers greater than million are $7$ digit numbers and completely divisible by 5. Now as the number is divisible by 5, two cases are possible:

1 When unit digit of the number is $0$:

$\quad \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {0}$

Number of ways rest of 6 digits can be arranged in 6 places $ = \dfrac{{6!}}{{3! \times 2!}}$ $= 60$

($\because $ Some digits (objects) are repeating)

2 When unit digit of the number is $5$:

Number of numbers which ends with $5$ = Numbers which end with 5 $-$ Numbers which starts by 0 and end with 5

Numbers end with $5$ $ = \dfrac{{6!}}{{3! \times 2!}}$ $= 60$

$\quad \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {5}$

Numbers which starts by 0 and end with 5 $ = \dfrac{{5!}}{{3! \times 2!}}$ $= 10$

$\quad \quad\underline {{0}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {5}$

Number of numbers which ends with $5$ $= 60 - 10 = 50$

Total ways $= 60 + 50 = 110$

We have 7 digits: $(2, 2), (3, 3, 3), 0, 5, 2$

We have to form the numbers greater than 1,000,000.

$\quad \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}$

Numbers greater than million are $7$ digit numbers and completely divisible by 5. Now as the number is divisible by 5, two cases are possible:

1 When unit digit of the number is $0$:

$\quad \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {0}$

Number of ways rest of 6 digits can be arranged in 6 places $ = \dfrac{{6!}}{{3! \times 2!}}$ $= 60$

($\because $ Some digits (objects) are repeating)

2 When unit digit of the number is $5$:

Number of numbers which ends with $5$ = Numbers which end with 5 $-$ Numbers which starts by 0 and end with 5

Numbers end with $5$ $ = \dfrac{{6!}}{{3! \times 2!}}$ $= 60$

$\quad \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {5}$

Numbers which starts by 0 and end with 5 $ = \dfrac{{5!}}{{3! \times 2!}}$ $= 10$

$\quad \quad\underline {{0}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {5}$

Number of numbers which ends with $5$ $= 60 - 10 = 50$

Total ways $= 60 + 50 = 110$

8 What is the sum of all five digit numbers formed by 2, 3, 4, 5, 6 without repetition?

A$20 × 5! × 11111$

B$20 × 3! × 11111$

C$20 × 4! × 11111$

D$20 × 4! × 10000$

Answer: C

Explanation:

Short cut: Use this formula: $(n-1)! × (1111...n\, times)$ $× (Sum\; of\; the\; digits)$

Here $n$ is the number of distinct digits given.

We have five digits 2, 3, 4, 5 and 6 we have to calculate the sum of all the five-digit numbers, which can be formed by using only these 5 digits without repetition. To do that we proceed in the following way:

If 2 is fixed at unit’s place then rest of the digits can be arranged on the remaining 4 places in 4! ways.

So, 2 appears 4! different numbers in the units place.

Similarly, remaining digits also appear in 4! different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times 4!$ $= 20 \times 4! \times 1$

If 2 is fixed at ten’s place then rest of the digits can be arranged on the remaining 4 places in 4! ways.

So, 2 appears in tens place in $4!$ different numbers.

Similarly, remaining digits also appear in 4! different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times 4! \times 10$ $= 20 \times 4! \times 10$

...

Similarly, Sum of all the numbers $= 20×4!×10000 +$ $20 × 4!× 1000$ $+ 20 × 4! × 100$$ + 20 × 4! × 100$ $+ 20 × 4! × 10$ $+ 20 × 4! × 1$

$= 20 × 4! × (10000 + 1000 + 100 + 10 + 1)$

$= 20 × 4! × 11111$

Short cut: Use this formula: $(n-1)! × (1111...n\, times)$ $× (Sum\; of\; the\; digits)$

Here $n$ is the number of distinct digits given.

*Detailed Explanation:*We have five digits 2, 3, 4, 5 and 6 we have to calculate the sum of all the five-digit numbers, which can be formed by using only these 5 digits without repetition. To do that we proceed in the following way:

If 2 is fixed at unit’s place then rest of the digits can be arranged on the remaining 4 places in 4! ways.

So, 2 appears 4! different numbers in the units place.

Similarly, remaining digits also appear in 4! different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times 4!$ $= 20 \times 4! \times 1$

If 2 is fixed at ten’s place then rest of the digits can be arranged on the remaining 4 places in 4! ways.

So, 2 appears in tens place in $4!$ different numbers.

Similarly, remaining digits also appear in 4! different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times 4! \times 10$ $= 20 \times 4! \times 10$

...

Similarly, Sum of all the numbers $= 20×4!×10000 +$ $20 × 4!× 1000$ $+ 20 × 4! × 100$$ + 20 × 4! × 100$ $+ 20 × 4! × 10$ $+ 20 × 4! × 1$

$= 20 × 4! × (10000 + 1000 + 100 + 10 + 1)$

$= 20 × 4! × 11111$

9 What is the sum of all five digit numbers formed by 2, 3, 4, 5, 6 with repetition?

A$20 × {3^5} × 11111$

B$20 × {4^4} × 11111$

C$20 × {4^5} × 11111$

D$20 × {5^5} × 11111$

Answer: C

Explanation:

Short cut: Use this formula: $(n-1)^4 × (1111...n\, times)$ $× (Sum\; of\; the\; digits)$

Here $n$ is the number of distinct digits given.

We have five digits 2, 3, 4, 5 and 6 we have to calculate the sum of all the five-digit numbers, which can be formed by using only these 5 digits with repetition. To do that we proceed in the following way:

If 2 is fixed at unit’s place then rest of places can be filled in ${4^5}$ ways.

So, 2 appears ${4^5}$ different numbers in the units place.

Similarly, remaining digits also appear in ${4^5}$ different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times {4^5}$ $= 20 \times {4^5}$

If 2 is fixed at ten’s place then rest of the digits can be arranged on the remaining 4 places in ${4^5}$ ways.

So, 2 appears in tens place in ${4^5}$ different numbers.

Similarly, remaining digits also appear in 4! different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times {4^5} \times 10$ $= 20 \times {4^5} \times 10$

...

Similarly, Sum of all the numbers $= 20×{4^5}×10000 +$ $20 × {4^5}× 1000$ $+ 20 × {4^5} × 100$$ + 20 × {4^5} × 100$ $+ 20 × {4^5} × 1$

$= 20 × {4^5} × (10000 + 1000 + 100 + 10 + 1)$

$= 20 × {4^5} × 11111$

Short cut: Use this formula: $(n-1)^4 × (1111...n\, times)$ $× (Sum\; of\; the\; digits)$

Here $n$ is the number of distinct digits given.

*Detailed Explanation:*We have five digits 2, 3, 4, 5 and 6 we have to calculate the sum of all the five-digit numbers, which can be formed by using only these 5 digits with repetition. To do that we proceed in the following way:

If 2 is fixed at unit’s place then rest of places can be filled in ${4^5}$ ways.

So, 2 appears ${4^5}$ different numbers in the units place.

Similarly, remaining digits also appear in ${4^5}$ different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times {4^5}$ $= 20 \times {4^5}$

If 2 is fixed at ten’s place then rest of the digits can be arranged on the remaining 4 places in ${4^5}$ ways.

So, 2 appears in tens place in ${4^5}$ different numbers.

Similarly, remaining digits also appear in 4! different numbers.

Sum of the digits in the units place = $(2 + 3 + 4 + 5 + 6) \times {4^5} \times 10$ $= 20 \times {4^5} \times 10$

...

Similarly, Sum of all the numbers $= 20×{4^5}×10000 +$ $20 × {4^5}× 1000$ $+ 20 × {4^5} × 100$$ + 20 × {4^5} × 100$ $+ 20 × {4^5} × 1$

$= 20 × {4^5} × (10000 + 1000 + 100 + 10 + 1)$

$= 20 × {4^5} × 11111$

10 Find the sum of all the five digit numbers formed by the digits 2, 4, 6, 8, 0 with out repetition.

A3199960

B4199960

C5199960

D6199960

Answer: C

Explanation:

(i) Use this following short cut:

$(n - 1)! × (111... n\; times)$ $× (Sum\; of\; the\;digits)$ $- (n – 2)!$ $× (111... (n - 1)\; times)$ $× (Sum\; of\; the\; digits)$

We have to make sure that $0$ should not be in the leftmost place.

So total numbers $-$ numbers end with $0$.

Therefore,

Sum of digits at unit’s place, ten’s place, hundred’s and thousand’s place $= (0 + 2 + 4 + 6 + 8) × (4! - 3!)$ $= 20 × 18$ $= 360$

Sum of the digits at ten thousand’s place $=(2 + 4 + 6 + 8) × 4!$ $= 480$

Total sum $= 480 × {10^4}$ $ + 360 × {10^3}$ $ + 360 × {10^2}$ $+ 360 × 10$ $+ 360$ $= 5199960$

(i) Use this following short cut:

$(n - 1)! × (111... n\; times)$ $× (Sum\; of\; the\;digits)$ $- (n – 2)!$ $× (111... (n - 1)\; times)$ $× (Sum\; of\; the\; digits)$

*Detailed Explanation:*We have to make sure that $0$ should not be in the leftmost place.

So total numbers $-$ numbers end with $0$.

Therefore,

Sum of digits at unit’s place, ten’s place, hundred’s and thousand’s place $= (0 + 2 + 4 + 6 + 8) × (4! - 3!)$ $= 20 × 18$ $= 360$

Sum of the digits at ten thousand’s place $=(2 + 4 + 6 + 8) × 4!$ $= 480$

Total sum $= 480 × {10^4}$ $ + 360 × {10^3}$ $ + 360 × {10^2}$ $+ 360 × 10$ $+ 360$ $= 5199960$

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