Question

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

A. 55
B. 66
C. 77  
D. 88
Answer :   77
Solution :
We have to form 7 digit numbers, using the digits 1, 2 and 3 only, such that the sum of the digits in a number = 10.
This can be done by taking 2, 2 , 2, 1, 1, 1, 1, or by taking 2,3,1,1,1,1,1.
∴ Number of ways $$ = \frac{{7!}}{{3!4!}} + \frac{{7!}}{{5!}} = 77.$$

Releted MCQ Question on
Algebra >> Permutation and Combination

Releted Question 1

$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$     and $$^n{C_{r + 1}} = 126,$$   then $$r$$ is:

A. 1
B. 2
C. 3
D. None of these.
Releted Question 2

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

A. 69760
B. 30240
C. 99748
D. none of these
Releted Question 3

The value of the expression $$^{47}{C_4} + \sum\limits_{j = 1}^5 {^{52 - j}{C_3}} $$    is equal to

A. $$^{47}{C_5}$$
B. $$^{52}{C_5}$$
C. $$^{52}{C_4}$$
D. none of these
Releted Question 4

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

A. $$^6{C_3} \times {\,^4}{C_2}$$
B. $$^4{P_2} \times {\,^4}{C_3}$$
C. $$^4{C_2} + {\,^4}{P_3}$$
D. none of these

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Permutation and Combination


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