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The number of $6$ digit positive integers whose sum of the digits is at least $52$ is

1. $21$
2. $22$
3. $27$
4. $28$

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Maximum possible sum of 6 digits is 54 when all the digits are 9.

answer is = sum is 52 + sum is 53 + sum is 54

sum is 54 = 1(all 9s)

sum is 53 =  one 8 and 5 9's = 6

sum is 52 = Four 9's and two 8's or five 9's and one 7 =

= $\binom{6}{4}$+ $\binom{6}{5}$ = 21

Ans = 1 + 6 + 21 = 28
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